15 - First properties of analytic functions
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welcome back to lecture five in our
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third week of the course analysis of a
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complex kind today we'll learn about
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some properties of analytic functions so
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far we have defined what an analytic
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function is by looking at what it means
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to be complex differentiable and then we
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looked at some examples and we defined
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the exponential function in
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trigonometric functions we looked at
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polynomials and so forth today we'll
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look at functions more generally and
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find some of their properties if we
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require that they are analytic here's a
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first theorem that we'll talk about
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today suppose F is an analytic function
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on a domain remember when a domain is
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that is an open and connected set this
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is all in one piece and suppose that you
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have that F prime of Z the derivative of
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F is equal to zero everywhere in D the
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theorem says that must imply that F is
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constant this is not all that surprising
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because we kind of know the one
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dimensional analog if F is a function
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that's defined on an integral and it's
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differentiable and its derivative is
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zero for all X and at Wisconsin and the
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image that we have in mind there is well
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what else could it be use my integral
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from A to B and I'm supposed to draw a
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function but the derivative is zero
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remember the derivative the
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one-dimensional derivative meant the
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slope of the tangent line so I'm
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supposed to draw a graph that is nice
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and in one piece and smooth and it has a
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tangent line with a zero derivative
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everywhere so the only way to do that is
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to draw something constant it was
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supposed to be constant so we'll use
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this one-dimensional fact and use it to
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prove the complex fact here's the idea
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of the proof first we show that the
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function f that's analytic and whose
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derivative is zero in all of T must be
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constant in any disk containing D so D
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is your domain and who knows what it
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looks like and I may have whole-new the
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it's not part of the this first fact
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sets as soon as you pick a disc that
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fits inside D in that disc at least we
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know that F must have the same value
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everywhere and the second part of the
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proof is then use that we already showed
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that F must be confidant any discs that
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we pick and the fact that D is connected
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to it's in one piece to show that F must
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be constant throughout D now let me show
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you how one would write these proofs if
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the first part of the proof suppose we
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have a disc that fits into D so here I
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just drew that disc without actually
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drawing D D is some domain that goes
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somewhere around we don't really know
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where it is but here's this disc that
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fits into D and Center we denoted by a
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and as radius is R it's really not
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important
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pick another point C in this disc an
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arbitrary point we're going to show that
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F at a and F at C must agree because C
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is an arbitrary point this will show
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that F must be constant it has to value
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FFA everywhere in this disk once you've
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picked a point C pick a point B as in
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this picture so that's on the same
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horizontal line at a and the same
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vertical line is C the derivative of F
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is 0 everywhere in D and if you write f
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as u plus IV then u X u Y VX and V Y
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must be 0 and E because F prime is U X
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plus I V X and you know the co should be
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many equations also give us that you
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wind v1 must be 0 so in particular let's
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look at the function U the real part of
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F on this line segment here on that line
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segment why doesn't change the imaginary
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part doesn't change only the
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x-coordinate changes so on this line
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segment the function U only depends on X
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and y is fixed and we have that the X
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derivative is 0 we can therefore look at
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you as a one-dimensional function here
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we fix y the 1
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it only depends on X and the X
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derivative is zero we can therefore use
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our one-dimensional fact and see that
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you must be constant on this particular
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line segment but that means that U of A
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must be equal to U of B well then you
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can do the same kind of arguing on this
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vertical line segment here the real part
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is fixing only the imaginary part varies
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so U is now a function of y only on this
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vertical line segment like the Y
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derivative is zero and again you can
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conclude from the one-dimensional fact
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that you up see and U of B must agree
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with each other but if u is the same at
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a at B and at C and must be the same at
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a and C so we have U of a is equal to U
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of C but C was an arbitrary point in
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this disc and therefore U is a constant
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function on this disc and you can see
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that similarly you could have made the
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same argument for V the imaginary part
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of F and also gotten that we must be
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constant on this disc so that shows the
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first part of the proof of a theorem
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here's the second part suppose a and B
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are now two arbitrary points in our
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domain D they don't necessarily lie in
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the same disc I can't draw a disc large
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enough around a that fits into D and
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contains B that disc would have to
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contain parts or not so I can't really
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use the first part directly to prove
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that the function must agree at a and B
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however D is connected and so I can find
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this nice curve here connecting a and B
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within D I could even make that from
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line segments if I wanted to by the
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previous step F is constant in a disc
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around a hydrologist that's small enough
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so it fits into D at Wisconsin then this
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green disc that I drew here but now
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let's look at this neighbor and orange
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disc that's another disc and in any disc
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at this constant so it's also constant
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on this
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disk but these two disks overlap the
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green and the orange disks they overlap
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on this piece here F is constant than
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the green disk so say F is equal to 1
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plus I on that bring this and it's also
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constant on the orange suit so that say
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if is equal to 2 there but on this blue
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portion where they overlap those two
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constants we're going to have to agree
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so the constant must be the same for the
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green disc and the orange disk so f is
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now constant on the union of the green
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and the orange disc what I can continue
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on like that look at the next is that
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overlaps of the orange disk and there's
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going to be constant on that next disc
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but that overlaps with the orange disk
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so has to be the same constant and so
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forth you continue on in this matter
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until I reach a disk that contains B and
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all together I can conclude that F of a
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and F of B must agree with each other
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and therefore at this constant in all of
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D
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now this theorem as simple as it is
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together with the Cauchy Riemann
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equations has very strong consequences
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here are some examples suppose again F
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is written as u plus IV and is an
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analytic function in a domain open
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connected set and suppose furthermore
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that the real part is constant this that
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implies that F must be constant in D you
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cannot find an analytic function whose
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real hard is constant and whose
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imaginary part varies as soon as the
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real part is constant
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everything is constant how would you
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prove that well you being constant
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indeed implies that the X and the y
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derivatives are 0 in D because use
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constant since F is analytic it
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satisfies the Koshi riemann equations
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and they imply that UX is equal to V Y
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since UX is 0v y must be 0 uy is equal
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to minus VX like uy 0 so BX is also
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equal to 0 so VX and V Y are also equal
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to 0 but now remember for an analytic
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function f prime is equal to UX plus IB
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X and we just showed that UX and VX 2
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both equal to 0 and that makes F prime
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equal to 0 but the previous theorem
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shows that F then must be constant so
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the Cauchy Riemann equations strongly
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correlate the real part and the
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imaginary part of an analytic function
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with each other they're not independent
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you cannot pick you to be constant and
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then have any choice on what to do with
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V once you pick you to be constant V has
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no more choices left
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similarly if F is an analytic function
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in a domain and V is constants of the
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imaginary part is constant then F must
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be constant and D as well also if f is
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analytic in a domain D again and its
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absolute value is constant remember the
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absolute value of f actually the square
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of the absolute value
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FZ can be found by taking the real part
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squared plus the imaginary part squared
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so if the absolute value of f is
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constant then this two implies that f
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itself must be constant so not even the
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absolute value can be constant so let me
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show you why here in the assumptions F
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is an analytic function in the domain
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and its absolute value this constant if
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the absolute value is constant in the
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square of the absolute value also can
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vary so therefore there must exist some
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constant C such that u squared of Z plus
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V squared of Z is equal to C no matter
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what Z you plug in right here that's
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what it means to be constant well
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there's two choices the C here can be
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equal to zero or cannot be equal to zero
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if it is equal to zero then this
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equation actually reads u squared plus V
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squared is equal to 0 but U and V are
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real valued functions the square of two
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real numbers is equal to 0 if and only
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if these two real numbers are themselves
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equal to 0 so this would mean that F had
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to equal to 0 and that's certainly
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constant and so in this case we're done
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with a proof because we have shown that
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F is constant constantly equal to 0 in
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fact if on the other hand this constancy
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is nonzero because U and V are real
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numbers and we're adding their squares
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the result has to be something positive
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so C is in fact positive now I'm sick
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trick we're going to look at this
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equation right here and we're going to
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take partial derivatives with respect to
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X and with respect to Y how do you find
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the derivative with respect to X of U
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squared of Z u squared of Z is the
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composition of two functions it's the
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function U of C composed with a function
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that squares the derivative of that by
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the chain rule is 2 times U of Z and
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then times the derivative of U itself
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which is UX because with
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the derivative with respect to X so the
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derivative of U square of Z with respect
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to X is 2 u u X similarly the derivative
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of V squared of Z with respect to X is
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to be the X and on the right hand side
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the derivative of a constant is 0 so
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this is the equation that results when
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we take a derivative with respect to X
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of this equation up here over here you
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see what happens when you take a
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derivative with respect to Y you get 2 u
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uy + 2 VB Y is equal to 0 now let's use
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the Cauchy Riemann equations remember
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that VX is equal to negative uy and
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we're going to substitute this fact in
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this first equation so instead of VX
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will write negative uy and in the second
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equation we're going to use that UX is
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equal to V Y and substitute V Y with UX
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and we also divide by 2 in here
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well we angled with is few UX minus v uy
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is equal to 0 and u uy plus vu X is
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equal to 0 now let's multiply this
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equation on the left by you while the
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Newton's do is u squared 10 2 X minus UV
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u I so equal to 0 and on the right we
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multiplied by the function v u v uy plus
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V squared UX is equal to 0 by the way
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I've been omitting the parameters Z u
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and parameter x and y because the
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equations would otherwise just look more
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complicated every single one of these
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equations is really u squared of X comma
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y plus UX of X comma y minus U of X
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comma Y times V of X comma Y x uy of X
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comma Y but it would make everything
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just look more complicated so these are
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the two equations that we were left with
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um last page now we're going to
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add these two equations what happens
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when I add these two equations to each
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other for the first equation we have a
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minus UV u I and in the second equation
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with a plus u bu what when we add the
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two left-hand sides to each other these
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two will cancel each other out
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and all that's left is the first term u
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squared u X and the second term of the
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second equation plus V squared Q X and
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the right-hand side zero plus zero add
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up to zero here we can factor out the u
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X so this reads u squared plus V squared
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u X is equal to zero but what is u
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squared plus V squared that was constant
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by our assumption that's our constant C
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so this last equation becomes C times u
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X is equal to 0 but C is a nonzero
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number in fact we showed it's positive
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the product of two numbers can only be 0
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one had the two factors of 0 which means
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u X must be 0 throughout D you can do a
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similar calculation by using different
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substitutions multiplying with different
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numbers and find that u wise also equal
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to 0 in D and now using the Cauchy
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Riemann equations once you have that you
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axe and you water it 0 indeed you can
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find that V X V while you're also 0
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indeed and therefore F prime is 0 and D
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and we're back to being able to use a
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theorem which says that F must be
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constant in D by the way D being
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connected is really important here
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because you could imagine a case where G
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consists of two open sets and you can
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have a function being constant equal to
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1 on this set and mean constant equal to
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2 on this set then the function overall
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wouldn't really be consonants only
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constant on these portions and so that's
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why D being connected it's important to
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imply that the function is constant
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throughout
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it's finished with a really strange
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example the strange example is the
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function f that's defined in the more
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complex plane by e to the minus 1 over Z
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to the fourth for nonzero Z values and
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zero at the origin clearly 1 over Z to
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the 4th is an analytic function analytic
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wherever Z is non zero so it's analytic
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in the complex plane minus the origin a
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minus sign still makes an analytic
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function the exponential function is
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analytic so the composition of these two
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is analytic so in C minus the origin
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clearly this function f is analytic in
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fact one can find you be there x and y
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derivatives and all that and show that
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they actually satisfy the Cauchy Riemann
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equations everywhere is they clearly
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satisfy them in C minus zero because we
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just saw that episode living in C minus
17:27
zero but they actually satisfy them at
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the origin as well and one can in fact
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show that the original UX uy VX and V Y
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are all equal to zero however it turns
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out F is not differentiable at the
17:44
origin how is that possible X satisfies
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the coachee riemann equations everywhere
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in c and is not differentiable isn't
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that a contradiction to some theorem
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that we had a revere in fact F isn't
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even continues at the origin so a
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function that's not continuous can't be
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differentiable remember if a function is
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differentiable then it's automatically
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continuous at that point since F is not
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continuous at the origin I'll show you
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why F cannot be differentiable at the
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origin so we'll have to solve this
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mystery
18:16
in a little bit so how is this possible
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for now I'm just going to show you why F
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is not continuous at the origin so
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you're the function again why is it not
18:26
continuous what is continuity mean again
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continuity means that the limit of
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F of Z in this case as the approach is 0
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should be equal to F of 0 that's what
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continuity with me the limit has to
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exist and as Z goes to 0 that lemon has
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to equal F of 0 so we need to look at
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what happens as Z approaches 0 what
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happens to F of C this'll limit exist
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and as always we're going to look at Z
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approaching the origin along some
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standard waves along the x axis along
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the y axis and so forth let's start Z
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approaching the origin along the real
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axis so Z is of the form X plus I times
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0 so no imaginary part and letting that
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go to 0 means X is going to go to 0 and
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F of Z is simply f of X and that's e to
19:19
the minus 1 over X to the 4th as X goes
19:22
to 0 this denominator here goes to 0 1
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over that goes to infinity but there's a
19:28
negative sign in front of it and so the
19:31
exponent here goes to negative infinity
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and e raised to a number that's going
19:37
toward negative infinity actually goes
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to 0 just remember the exponential
19:41
function looks like this and if you plug
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in numbers that are closer and closer to
19:47
negative infinity the exponential
19:49
function gets very close to zero so as X
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approaches the origin along the real
19:56
axis f of X goes to 0 all right so maybe
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something will happen along the
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imaginary axis so that's next consider Z
20:04
approaching the origin along the
20:06
imaginary axis that means Z is off the
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form iy no real part just iy and that's
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going to go to 0 so Y goes to 0 what's F
20:14
of Z what's then F of I Y which is e to
20:18
the minus 1 over I to the fourth Y to
20:21
the fourth but I to the fourth that's
20:23
just 1 so it's e to the minus 1 over Y
20:26
to the fourth and the exact same
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reasoning that we used before shows and
20:30
again F goes to 0 so f goes to 0 as you
20:34
approach 0 both along the real and the
20:38
imaginary axis so so far so good Rick
20:40
where's the problem so far it looks like
20:42
F is indeed continuous however consider
20:47
our Z at the forum our e to the I PI
20:51
over 4 so Z's going to 0 along a
20:57
45-degree angle what is Z to the fourth
21:02
in that case once R to the fourth times
21:05
e to the I PI over 4 times 4 which is e
21:08
to the I pi and what is e to the I PI e
21:12
to the I PI is negative 1
21:15
so Z to the fourth is minus R to the
21:19
fourth and that minus sign is what's
21:21
going to be the cause of all trouble
21:23
what's F of Z it's even the minus 1 over
21:27
minus R to the fourth and these two
21:29
minus signs will cancel each other out
21:31
so F of Z is now the 1 over R to the
21:34
fourth but as R goes to 0 1 over R to
21:39
the fourth goes to infinity there's no
21:41
negative sign layer to rescue us the
21:44
exponential function asks you plug in
21:46
numbers that are bigger and bigger
21:47
towards plus infinity goes towards plus
21:50
infinity and that is certainly not equal
21:53
to f of 0 so the limit doesn't even
21:56
exist and it certainly doesn't equal F
21:59
of 0 it shows that the function is not
22:01
continuous at 0 so with F knocking
22:06
continuous at 0 it's also not
22:08
differentiable at 0 so let's go back to
22:10
the original problem what happened you
22:13
and we satisfy the Cauchy Riemann
22:15
equations everywhere and see yet ever
22:18
it's not differentiable at the origin
22:19
how is that possible well this is what
22:22
if they are really set let F be defined
22:26
on the domain D then f is analytic
22:31
indeed if and only if you and we have
22:36
continuous first partial derivatives in
22:39
D that satisfy the Cauchy Riemann
22:41
equations so it's not quite enough just
22:44
to satisfy the Cauchy Riemann equations
22:46
the partial derivatives have to be
22:48
continuous
22:54
in our strange example these partial
22:57
derivatives are not continuous at zero
23:00
and so the assumptions of the theorem
23:02
are simply not satisfied even though
23:04
they seem to satisfy the Cauchy Riemann
23:06
equations that's not enough they need to
23:08
be continuous
23:10
in addition so that's what went wrong in
23:13
this example it's a very subtle example
23:16
and most of the time this won't affect
23:19
you in the least
23:20
next week we'll study conformal mappings
23:23
those are analytic functions that are
23:26
injective or one-to-one
23:28
in addition and now have some very
23:30
special properties that we'll learn
23:31
about we'll talk about some examples of
23:34
conformal mappings namely the Mobius
23:36
transformations that are fascinating in
23:39
and out themselves and as a bonus i'll
23:42
show you the riemann mapping theorem
23:44
which says the following suppose i'm
23:47
giving you some kind of domain again
23:50
open connected set but this time i'm not
23:52
allowing that domain to have a whole
23:54
journey domain that's not having a whole
23:57
so something like this it's not allowed
24:02
so the only kind of domains that I'm
24:04
going to allow are the ones that have no
24:06
holes suppose I'm giving you such a
24:09
domain then I can find an analytic
24:13
function that is wonderful and and on to
24:19
from this domain this no matter how ugly
24:22
you drew that on to the unit disk and
24:25
it's going to be a conformal mapping so
24:27
it's going to be analytic one-to-one
24:33
injective I'll explain what that means
24:34
next week and on to so in other words
24:38
the riemann mapping theorem makes it
24:40
possible to transfer things back and
24:43
forth between this domain and thinking a
24:46
disk this is one of the most amazing
24:48
theorems and complex analysis we will
24:50
learn about it next week