Roots of complex numbers and exponentials
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Finding roots of complex numbers is geometrically much richer, even if we can only accomplish such approximately. In this lecture we show how exponential polar coordinates reveals the pleasant 2D geometry involved, in in the special case when we want to solve zn=1. The cases n=1,2,3,4 and 5 are all explicitly dealt with, and we try to illustrate the geometrical significance of the algebra by playing with GeoGebra.
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hello everyone I'm Norman wah Berger in
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this lecture we're going to be looking
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at roots of complex numbers so this is
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touching base with a very rich theory of
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polynomials and their solutions when we
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go to the complex plane we get a richer
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and more satisfying picture to the
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algebra of polynomials so a good place
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to start is with the simpler story of
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finding roots of complex numbers what do
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we mean by that we mean solving an
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equation whoops solving an equation that
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looks like this
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that looks like Z to the say N equals
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some complex number W so W is a complex
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number and we want to find solutions of
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this equation Z to the N equals W and
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typically N is a natural number so that
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is a much richer story than it is over
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the usual number line over the so-called
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real numbers and it will turn out that
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typically we get actually n distinct
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solutions to such an equation and a good
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place to start is with a very special
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but important case when the W is
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actually 1 so the special case is when
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we considering equations of the form Z
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to the N equals 1 solutions of this
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equation are called roots of unity
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solutions are more precisely nth roots
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of unity and roots of unity unity being
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one
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okay so let's have a look at this
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equation for simple values of n that's
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an instructive place to start
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so what N equals 1 well then it's a
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pretty simple equation Z equals 1 and
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there's only one solution to that which
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is Z equals 1 so on the complex plane
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I might actually draw our unit circle
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which we like a lot ok and so the
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solution here is just considering just
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that point there
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ok how about when N equals 2 so now
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we're talking about Z squared equals 1
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ok there's two solutions to that and
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they are Z equals plus or minus 1 and a
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reason why we can see that is because Z
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squared minus 1 when we bring everything
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to one side that factors as Z plus 1
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times Z minus 1 ok so it's really the
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factorization of this into these linear
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factors that gives us the the roots or
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the zeros and there are two of them in
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this case plus and minus one so if we do
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a picture maybe I'll draw it up here the
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picture for this second case would be
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there's this one and this one ok how
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about when N equals 3
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now we're talking about Z cubed equals 1
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now that's a little bit trickier so how
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would we solve that well if we wanted to
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look at high school kind of approach we
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might think about trying to factor it
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and one factor is pretty easy we can
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extract as Z minus 1 from that and then
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the remaining factor is Z squared plus Z
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plus 1 ok
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which you can verify by just expanding
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out and it's kind of obvious that Z
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minus 1 is a factor because obviously
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this thing here has a 0 of 1 so that
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minus 1 is a factor but there's a this
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additional quadratic factor so to find
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the other solutions to this thing what
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we should do is we should try to see
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when this second quadratic factor is
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also 0 so let's solve
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this quadratic equation Z squared plus Z
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plus 1 equals 0
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okay that's a quadratic equation we know
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that well ordinary quadratic equations
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have the quadratic formula recipe for
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solutions and in fact sort of
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algebraically that works with complex
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numbers to subject to a particular
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little issue which we'll we'll see in a
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second
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ok so let's apply the quadratic formula
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what would that tell us we'll say that Z
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is going to be equal to so minus B since
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I get some minus 1 and then plus or
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minus square root of B squared ok 1
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minus 4ac so that's about minus 4 all
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over 2 and what is that well that's
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equal to minus 1 plus or minus the
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square root of minus 3 over 2 so to find
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a solution to this we need we need a
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square root of minus 3 now in high
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school there is no such number and we
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would say that this quadratic equation
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doesn't have a solution but in the
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complex plane we have a little bit more
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flexibility because we have I and we
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know that I squared is equal to minus 1
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so we can interpret this square root of
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minus 3 to be I times the square root of
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3 or minus I times the square root of 3
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so if we have a square root of 3 like a
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real number then we multiply that by I
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or by minus I and we square that then
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we're going to get minus 3 okay so
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there's two solutions here so to come
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back to our equation here we could write
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this now as minus 1 plus or minus I
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times root 3 on 2 ok so where are those
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numbers now let's have a look at then
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three cube roots of unity okay there's
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one there's I so one solution is this
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one here which is one and another
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solution is right here okay where we
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have to go minus 1/2 minus 1/2 there and
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up root 3 on 2 I and so this is minus 1
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plus I root 3 all over 2 and the other
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one where we have a minus sign rather is
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down here so that'll be minus 1 minus I
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root 3 on 2 and we know from our
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familiar experience with trig things
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that we have a situation here where
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there's a kind of a 60 degrees this is
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60 degrees or PI by 3 so this is really
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a third of the way around so this is 2
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pi on 3 and this is also 2 pi on 3 and
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this is also 2 pi 3 in other words these
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three points that we have make a nice
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equilateral triangle which is embedded
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in this circle ok so finding the 3 roots
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3/3 roots of unity is essentially
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geometrically same as finding this
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equilateral triangle in the unit circle
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where one of the vertices is 1 so to
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summarize what we've done we have found
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therefore that the solutions solutions
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to Z cubed equals 1 are ok this Z equals
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1 their Z equals minus 1 plus I root 3
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on 2 and there's Z equals minus 1 minus
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I root 3 on 2 and these form
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an equilateral triangle in the unit
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circle excellent okay great
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so what about the next one let's go to N
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equals four
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now we're talking about Z to the 4 minus
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1 well we could start by factoring that
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that's a difference of squares so it'll
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be Z squared plus 1 times Z squared
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minus 1 and okay can we factor that well
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one of the factors is easy to factor we
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get Z minus 1 times Z plus 1 and in high
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school you wouldn't be able to factor
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the Z squared plus 1 because it doesn't
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have any zeros in the real numbers but
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with the complex numbers we can't factor
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that because this is looking for
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solutions of Z squared equals minus 1
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and we know the solutions are I and
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minus R so we can in fact factor this as
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Z - I said plus I said minus 1 Z plus 1
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and so the 4th roots of unity for 4th
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roots of unity are Z equals I'll just
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lift 1-1 I and minus I and geometrically
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that's a very nice situation one of them
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is there that's one and then I is up
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here and minus one is here and minus I
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is down here so again we are getting a
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nice regular polygon this case a square
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inscribed in the unit circle ok so
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that's great so we can kind of perhaps
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expect that something like that is going
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to happen for higher powers and that's
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exactly what happens in general in
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general
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the solutions to Z to the N equals one
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form a regular n-gon
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on the unit circle and wait to see that
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is - let's let Z be say our e to the I
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theta that's a general polar form for a
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complex number and we'll stick that into
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the equation then we get r times e to
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the I theta to the nth power equals one
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okay so what's that amount - okay so or
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this is R to the N times e to the I and
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theta equals one now what's the polar
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form for one well one is if we write it
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in polar form it's 1 times e to the I
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times 0 really right because the angle
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is 0 the argument is 0 so if we compare
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these two sides of the equation here we
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kind of get two equations one of them is
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that R to the N equals 1 and the other
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one is that e to the I and theta equals
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0 okay what are the solutions to that
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well R is a positive number a real
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number so there's only one solution to
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that R has to be equal to 1 and what
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about e to the I and theta equals 0
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well if you don't make it easy I intend
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that would suggest that n times theta is
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equal to either 0 or maybe 2 pi or maybe
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4 PI etc okay and a way to get all of
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those possibilities at the same time
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is to write it rather as n times theta
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equals to PI K for some integer actually
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not just a natural number integer K so
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in said okay so that is the requirement
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to remember that the exponential has a
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periodic aspect if we increase the
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argument by two pi then we go back to
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where we're starting because we're going
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around the unit circle so this second
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equation here will be valid when n theta
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is some multiple of two pi okay and that
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tells us then that theta is going to be
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two pi over n times K in other words a
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multiple of two pi over n so when we put
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all of that together what do we see we
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end up with a situation where we have
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said having to be on this unit circle
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and one solution would be that is e to
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the I 2 pi on n another solution
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well actually one solution would be this
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one these necessarily actually the
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simplest one that's e to the I 0 0 times
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2 pi on it okay and then we have this
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one and then the next one over would be
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e to the I 4 pi on n twice the argument
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of this of this one doing the same angle
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and then the next one would be e to the
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I 6 pi on N and we go around like this
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and there's going to be altogether
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n solutions because each one of these
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angles is 2 pi over n 2 pi over n 2 PI
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over N
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so the last one will be e e to the I to
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PI over n times n minus one
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after that we're back to this one here
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so this is the same as e to the I to pi
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over n times n ok we get back to where
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we started so we end up getting a an end
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gone a lot of equally spaced points all
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the way around making a regular n-gon so
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this entire problem connects with a very
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old and famous problem of sort of how to
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construct a regular and gone ok so that
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geometrical problem is intimately
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connected with this arithmetic 'el 1 or
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algebraic 1 when we're talking about
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complex numbers so I want to show you
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change gears here a little bit what
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happens here in the unit circle I'm
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gonna illustrate things with geogebra so
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I have the unit circle here and I've
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started with this point as ed 1 ok and
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maybe just make that slightly different
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color so we can see it better maybe blue
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ok so there's that one now what we see
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here in addition to that one is all
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kinds of sorry about that ok there we go
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so if there's that one then this Z 2 is
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Z 1 squared is the square then this is
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the cube and then this is the fourth
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power so these a successive powers of
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this is at 1 and here are negative power
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so here is it said 1 to the minus 1 Zed
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1 to the minus 2 and so on so I want you
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to have a look at what happens to these
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powers as I move Zed one around one of
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the beautiful things about geogebra is
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were allowed to do this and it will just
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move everything accordingly so notice
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that these positive powers are just
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marching around
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okay so whoops okay so suppose like I
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brought it all the way up to here so I'm
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just said one is now at I what's
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happened is that all those higher powers
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and the negative powers also they all
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coincide with these fourth roots of
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unity so this is Z 1 this is at one
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squared
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this is at 1 cubed as I'd went to the
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4th Zed 1 to the 5th
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Zed 1 to the sixth and so on and going
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backwards this is one this is said 1 to
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the minus 1 this is it went to the minus
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2 and so on now if I carry on a little
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bit say move it close to about 120
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degrees then there we go we see the 3/3
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roots of unity appearing so all the
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successive powers are rotating around in
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this triangular fashion and whoops we
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could go back another kind of
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interesting case is there's at some
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point here I should get a regular 5 gone
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so the rate there you can see that we
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have five points so this is the story of
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the fifth roots of unity which is a
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little bit algebraically more tricky to
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establish so let's have a look at that's
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actually quite an interesting situation
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ok so let's now go back to our
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our screen yeah okay so now let's have a
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look at N equals 5 where we're talking
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about the solutions of Z to the fifth
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equals one so if we want to factor this
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the way we do that as we take that minus
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one and then we get say Z to the fourth
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plus Z cubed plus said squared plus said
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plus 1 now factoring this quartic is
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let's say a little bit trickier so here
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is one way of thinking about that so if
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we take Z to the fourth both said cubed
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plus Z squared plus Z plus one here's a
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little bit of a trick if we write this
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as Z squared plus one-half Z plus one
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all squared if you square this thing
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here you'll see you get almost this
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thing you just have to make a little
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compensation you have to actually put in
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minus five quarters of Z squared to make
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everything equal so you can check this
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alright you can check this little
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identity and once you've done that you
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see that what you have on the right hand
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side is a difference of squares so it's
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a difference of squares that means we
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can write it as product of two linear
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factors and one of those linear factors
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gonna be well it's gonna be this first
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one plus the square root of this thing
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and then the other one is going to be
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this first one minus the square root of
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this thing so we write down those linear
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factors and we're gonna get Z squared
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plus 1 minus root 5 on 2 times Z plus 1
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and the other one is going to be Z
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squared plus 1 plus root 5 on 2 times Z
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plus 1 so we get a kind of a non-obvious
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factoring of this kortek into these two
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quadratic factors but the quadratic
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factors themselves have these
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Gerudo fives in them okay so that means
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if we want to completely factor this
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thing out well then we would have to
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essentially apply the quadratic equation
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to both of these quadratic factors and
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we could do that and I'm gonna leave
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that as an exercise and what do we end
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up getting so we get the fifth roots of
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one I'm gonna show you the picture okay
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so here is our unit circle and okay
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there's one here this one here the one
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here there's one here and all together
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these should form a regular pentagon
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okay regular pentagon and it's a little
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bit complicated what the actual points
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are here but let me write down what they
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are okay so this one here turns out to
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be root 5 minus 1 over 4 that's the real
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part plus the square root of 10 plus 2
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root 5 inside the bigger square root all
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over 4 times I and this one down here
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well it's gonna be the complex
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conjugates okay look if that's said then
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this one here is going to be use ad bar
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you'd obtain that one just by changing
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this plus sign in here to a minus sign
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and what about this one over here well
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it's similar but a little bit different
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it's minus root 5 minus 1 over 4 plus
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square root of 10 minus 2 root 5 over 4
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times I and this one here is the complex
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conjugate of that
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and these are the solutions that you get
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by solving the two quadratic equations
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that are appearing here and once you
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have done this then what we get is we
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get as a consequence we find the values
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of cosine cosine 2 pi on 5 right this
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angle here is 2 pi on 5 and so the
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x-coordinate this thing here is going to
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be cosine of 2 pi + 5 and the
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y-coordinate is going to be sine of 2 pi
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on 5 that's going to be with that thing
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there and of course we could also go a
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little bit further and if we wanted to
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find this angle here that's going to be
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4 PI on 5 so we can also get values of
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cosine of 4 PI on 5 + sine of 4 PI on 5
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from these they're they're going to be
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these values here and here now this is
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very explicit this is in Cartesian
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coordinates we've used some algebraic
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trickery in order to factor these things
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from the point of view of complex
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numbers another way of describing these
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things is in terms of polar coordinates
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and basically in other words we can
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connect now with the previous formula
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where I told you in general what you get
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up here the general formula will be in
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terms of multiples of 2 pi over 5 so
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maybe I'll do it in a different color
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how about about okay so this number here
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has an alternate expression as e to the
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I 2 pi on 5 because that's the argument
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right there that's e to the I to point 5
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and this one here is e to the I 4 pi on
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5 and this one here is
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well e to the minus I 2 pi on 5 and this
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one here is e to the minus I 4 pi on 5
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so that's another way of describing
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these these fifth roots of one ok so
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once we have this sort of technology in
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place then actually we can extend this
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to get nth roots of more general complex
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numbers ok so suppose now we want to
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solve let's say solve Z cubed equals 5 I
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okay so here's our unit circle one I and
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up here somewhere is 5i all right now we
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want to find complex numbers whose cube
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is this point 5i so now we're looking
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for a point Zed whose cube is equal to
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this point 5i up here but we're no
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longer on the unit circle so we have to
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be a little bit more general so to
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prepare us for that let's have another
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look at geogebra here I remind you
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there's the point Z 1 and here the
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various powers now I'm going to
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disassociate the point from the circle I
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have a look at what happens to the
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powers so they still march around at an
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angular regular fashion but they're
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moduli are increasing and that's because
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the modulus of the square is the square
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of the modules so if this thing here has
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modulus a one point one then this next
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power will have modulus 1 point 1
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squared and this will have modulus 1
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point 1 cubed and the inverse will have
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modulus 1 over 1 point 1 and this will
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have modulus 1 over 1 point 1 squared
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and so on and that becomes of course
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more extreme as we get further and
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further from the
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unit circle so if we make everything a
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bit smaller so we can see more clearly
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okay so actually let's see if we can
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find our point five so I'll write down
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five I hear some mercy if we can get
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that point five I okay there's five I up
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here it's Z 12 so we're looking to put Z
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in a such a place so that this third
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power Z sub three coincides with this
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point five I so let's see if we just do
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that on the page here okay there so
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getting pretty close so how you do it
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exactly
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okay the row there thought bad so here
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is a value of Z for which here is N
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squared and here is Z cubed so this
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would be one at least approximate
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solution to the equation Z cubed equals
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five I now I want to show you where the
28:43
other solutions are to do that let me
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draw a circle circle with Center through
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point okay here's I'm going to draw a
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circle now centered here through this
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point here okay now I'm going to take
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this point Zed one I'm going to rotate
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it by 1/3 of a full turn so let's see
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what happens as I do that I'm gonna go
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around whoops don't go back sorry let's
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take the point I'm going to try to move
29:11
its roughly on this circle okay
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whoops I'm moving it around this circle
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roughly
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okay can you see right about now
29:33
somewhere around here okay it's going to
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be another value for which the square
29:40
now is down here and the cube is up here
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pretty close to five I again so what's
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going to happen is there are going to be
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three solutions and they'll be the one
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that I showed you first and then there's
29:51
this one over here and there will be
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another solution down here and it'll
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turn out that these three solutions
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themselves form a nice little
29:59
equilateral triangle on this bigger
30:01
circle okay so let's see if we can see
30:04
that algebraically okay so we'll go back
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here and we are going to approach this
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with polar coordinates
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okay so let's use polar coordinates so
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we use polar coordinates namely we let Z
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be our e to the I theta and we also
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write 5i also in polar coordinates so
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the modulus of 5i is 5 and the angle is
30:44
PI by 2 so it's e to the I PI by 2 okay
30:52
so now we have then that Z cubed equals
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5i becomes the equation our e to the I
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theta cubed equals 5 e to the I PI on 2
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or R cubed times e to the 3 I theta
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equals 5 e to the I PI on 2
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so what can we deduce from that so from
31:27
that we deduced that the moduli of both
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sides of the same so R cubed equals 5
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and terms of the arguments 3 theta has
31:42
to be equal to PI by 2
31:45
up to a multiple of 2pi and it's crucial
31:48
that we allow this flexibility in the
31:52
angle by introducing plus 2 pi K where K
31:56
is an integer which will write like this
31:59
ok so there's actually quite a few
32:01
possibilities encoded in this second
32:04
equation so the first equation because R
32:07
is a positive real number has to be R
32:10
equals cubed root of 5 so we write down
32:14
that symbol of course that's in general
32:16
something that we can only calculate
32:18
approximately but nevertheless it has a
32:21
unique value and what about theta
32:24
well solving for theta we get theta
32:26
equals PI over 6 plus 2 pi on 3 times K
32:34
so there are now going to be three
32:37
solutions because as K varies through
32:40
the integers this 2 pi on 3 will repeat
32:44
itself every 2 pi whenever K goes up by
32:48
by 3 so we have get solutions get
32:56
solutions theta equals PI over 6 so
32:59
that's when K equals 0 when K equals 1
33:04
we're gonna get theta equals PI over 6
33:07
plus 2 pi on 3 which I suppose we could
33:11
rewrite as in terms of 6 that's PI over
33:15
6 plus 4 PI on 6 all together 5 PI on 6
33:20
so that's when K equals 1 and when K
33:25
equals 2 we would get the PI over 6 plus
33:28
4 PI on 3 which equals that's 8 pi on 3
33:38
so plus 1 so it's 9 PI on 6 9 PI on 6
33:45
and that's when K equals 2 and let's
33:51
just have a look what would happen if we
33:53
went up to K equals 3 so for k equals 3
33:58
we would get theta equals PI over 6 plus
34:02
2 pi okay
34:05
but that's really the same as PI over 6
34:07
so that means we're just repeating the
34:10
value that we got initially here right
34:12
so that's what happens after every 3 we
34:15
repeat the values that we already have
34:17
so we're really getting three solutions
34:18
PI over 6 5 PI on 6 and 9 PI on 6 okay
34:25
so to summarize so therefore the
34:27
solutions are Z equals the cubed root of
34:36
5 times e to the I PI on 6 the cubed
34:42
root of 5 times e to the I 5 PI on 6 and
34:47
the cubed root of 5 times e to the I 9
34:52
PI on 6 ok
34:55
so these could be perhaps simplified we
34:57
do know that PI over 6 is the same as 30
35:00
degrees so we could write this in
35:02
Cartesian coordinates if we want to do
35:04
but it's kind of useful to just regard
35:09
it like this and geometrically what are
35:13
we getting
35:13
there's our unit circle the cubed root
35:16
of 5 is a little bit bigger than 1 but
35:20
not not as biggest 2 so maybe here is R
35:25
equals cube root of 5 and then our three
35:31
points are ones where the modulus is PI
35:34
over 6 so that's like this one over here
35:38
ok and then the next one is 5 PI on 6
35:42
it's gonna be something like this and
35:44
then 9 PI on 6 it's the same as 3 over 2
35:49
that's gonna be actually right down here
35:50
so these are the three solutions to Z to
35:56
the cubed equals 5 I geometrically
36:00
and here are the corresponding algebraic
36:02
values all right so good we've had a
36:09
look at nth roots of unity and both a
36:13
geometrical and an algebraic approach
36:15
using polar coordinates and relying very
36:18
much on the exponential polar form and
36:20
we've been able to extend this to taking
36:25
nth roots of more general complex
36:27
numbers this is all in the direction of
36:31
solving more complicated polynomial
36:34
equations in the complex domain so I'll
36:38
see you next time