34 - Finding Residues
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welcome to the fourth lecture in our
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seventh week of our course analysis of a
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complex kind this lecture will focus on
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new tricks to find residues now that we
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have the powerful residue theorem we
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need new techniques to find residues
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more easily let me remind you of the
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residue theorem suppose f it's a
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function that is analytic except for an
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isolated singularity so that means F is
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analytic and a punctured disk centered
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at 0 with the exception of Caesar in
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that case FS of Laurent series
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representation in that punctured disk
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which is a doubly infinite series of
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terms a cadiz of the coefficients times
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e minus e0 to the cake Ricky runs from
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negative infinity to positive infinity
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we see the residue of F at to 0 it's the
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a minus 1 term this is the term for
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which K is equal to minus 1 and that is
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the coefficient of the 1 over Z minus 2
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0 term let's start by finding residues
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at removable singularities recall that a
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removable singularity is a singularity
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for which you find out that after
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finding the Laurent series there are
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actually no terms that have negative
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powers of C minus easier so all the AKS
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are actually equal to 0 for K less than
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zero one particular than n minus 1 must
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be equal to 0 which is the residue so in
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that case the residue of F at C 0 is 0
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let's look at an example the come to
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sine Z over Z sanity are busy as an
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isotope at singularity at the origin
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because that's what we're dividing by 0
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otherwise the function is analytic sine
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Z itself has a Taylor series
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representation centered at the origin
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namely z minus z cubed over 3 factorial
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is Z to the fifth over 5 factorial and
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so forth what you see right here if I
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divide that by Z I find myself with 1
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minus 1 over 3 factorial times Z squared
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plus 1 over 5 factorial times e to the
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4th and so forth that's the LeBron
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series representation of signs you
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receive which we use to show that Z
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equals 0 is actually a removable
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singularity for this function f
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therefore the residue
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zero is equal to zero next let's
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consider residues at simple pulse
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recall that c0 is a simple pole if the
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eight minus one term in the Laurent
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series representation is nonzero but
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other Akkad's for case negative 1 are
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equal to zero so there's exactly one
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negative power of C minus a 0 in the
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Laurent series expansion so f has the
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form a minus 1 divided by C minus C 0
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and then all non-negative powers of Z
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minus u sir how do we find this key
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minus one term he is the idea we
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multiply through by Z minus C 0 we
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multiply both sides of the equation by Z
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minus 2 0 we found that D minus C 0
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times f of Z is Z minus 2 0 times a
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minus 1 over Z minus a 0 so that's just
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a minus 1 and then each term gets an
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additional t minus T 0 so a 0 times Z
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minus 2 0 a 1 times Z minus a 0 squared
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and so forth now if I'm at Z approach to
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0 and take the limit then as Z
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approaches to 0 this term goes away this
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term goes away and all I'm left with is
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this term right here so the residue of
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Evans is here can be found by taking the
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limit as the approaches C 0 of Z minus C
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0 times F of C let's see if we can use
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that in an example
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let's look at F of C equals one over Z
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squared plus one we've already looked at
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this function before we know it has
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simple poles it's a 0 equal to I and 2 0
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equals negative I will focus on I right
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now and try to find the residue and c0
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equals on by our formula you can find
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that residue by first multiplying the
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function f by Z minus I that's what you
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see right here multiply F by Z minus I
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and then you need to take the limit as Z
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approaches I but 1 over Z squared plus 1
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is actually equal to 1 over Z minus I
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times Z plus 9 because we can factor the
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denominator the Z minus I in the
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denominator cancels authenticity - I so
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that we're left with simply 1 over Z
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plus I the limit as Z approaches I off
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that fraction can be found easily by
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simply plugging in I for Zeke and if I
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apply an eye for C I get 1 over 2i and 1
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over 2 is the same thing as negative
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vibration so the residue of this
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function Phi is negative PI over 2
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that's the same residue we found last
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class with a very different method
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look at double polls a double pole is
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one for which the a minus two term is
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nonzero but all further in Cades for K
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less and we go to negative three or
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equal to 0 so f has the form e minus 2
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or 1 Z minus is 0 squared then maybe
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there's an e minus 1 term or no and then
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all the rest of the terms are
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non-negative powers of C minus 2 0 how
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do we isolate a negative 1 the idea is
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very similar
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this time we multiplied through by Z
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minus Z 0 squared the largest
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denominator we have so C minus C 0
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square root x fo C gives us it's simply
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minus 2 and the a minus 1 term gets an
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extra as u minus C 0 a 0 gets as u minus
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a 0 squared and so forth we can't simply
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let the approach to 0 because if we did
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that we would isolate a negative 2 we
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don't interested a negative 2 we're
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interested in a negative 1 that's why we
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have an additional trick right here we
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take a derivative if we take a
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derivative of this equation then the a
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minus 2 term goes away
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we're left with an a minus 1 term the
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derivative of a zero times u minus C 0
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square this to a zero tendency - easier
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and all the subsequent terms have some
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power of C minus a 0 now if he let the
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approaches easier
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all these subsequent terms go away and
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were left with a negative 1 so we found
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our formula for double poles the residue
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of the double pole can be found by first
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multiplying the function by Z minus c0
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is squared and then taking a derivative
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and then taking the limit as Z
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approaches to 0 let's look at an example
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this up at f of z equals 1 over z minus
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1 square Khem c minus 3 that function
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has a double pole and c0 equals 1 it
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also has a simple pole at 3 but we're
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not interested in that Pole right now we
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won't define the residue of F + 1 by our
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formula we need to multiply this
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function by Z minus 1 squared
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that's what's happening right here then
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need to take a derivative and finally
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the limit as Z approaches one we see
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right away multiplying by the minus 1
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squared is very convenient because the Z
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minus 1 squared term will simply cancel
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out so that we're left with having to
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find the derivative of 1 over Z minus 3
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and then the limit as C approaches 1 but
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the derivative of 1 over Z minus 3 is
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minus 1 or Z minus 3 squared and the
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limit of that as Z approaches 1 we
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simply have to plug in 1 for Z 1 minus 3
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is negative 2 negative 2 is where's 4
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and so that the residue of our function
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and 1 is negative 1/4
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let's try to generalize these ideas
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suppose now we're dealing with a pole of
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order n that means the a minus n term is
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nonzero and all other aks for K less
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than or equal to negative n plus 1 those
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are equal to 0 so my function looks like
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this I have a minus n over Z minus 0 to
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the N and then terms with a denominator
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gets better and better all the way
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through a minus 1 over Z minus C 0 and
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then the non negative powers of C minus
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easier the idea is the same thing we're
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going to multiply through by Z minus C 0
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to the N and if I do that I'm left with
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a minus n plus a minus n plus 1 times C
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minus C 0 all the way through a minus 1
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times C minus 0 to the N minus 1 plus a
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0 10 C with a 0 to the N and so forth I
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want to isolate the Eamonn is one term
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so I can't simply let Z approaches to 0
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I need to isolate this term and
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therefore I need to take an N minus 1
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folder if I take a derivative n minus 1
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times this D minus C 0 to the N minus 1
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will go away all these terms in the
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front will also go away under
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differentiation and only terms to the
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right will be left so if I differentiate
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this n minus 1 times I find the
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following the first time I take the
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derivative off the a minus 1 term I get
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an N minus 1 from the exponent then a
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minus 1 times Z minus Z 0 to the N minus
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2 the next time I differentiate to get
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the factor n minus 2 up front and so
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forth if I differentiate n minus 1 times
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the Z minus c0 term has gone entirely
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and I have an N minus 1 times n minus 2
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times n minus 3 and so forth all the way
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down to 1 times and when this one term
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left the a 0 term also has all kinds of
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things up front here but it has a Z
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minus a 0 term left and so do all the
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subsequent turns now if I let Z approach
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C 0 well all these subsequent terms will
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go away so if I now take a limit as Z
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approaches C 0
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then I can isolate my I'm in this one
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term if I simply divide by this number
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right here which happens to be n minus 1
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factorial so we just found the following
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formula the residue of F at a pole of
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order n is 1 over n minus 1 factorial
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times the limit as C approaches C 0 the
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enma is one full derivative of Z minus
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is 0 to the N times F of C this formula
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clearly contains the two cases that
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we've already considered earlier the
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simple pool and the double pool but now
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you can use it for higher order poles as
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well to finish I want to show you one
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other very useful little fact suppose
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you have a function f that is a quotient
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of two analytic functions G and H where
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H the denominator has a simple zero at
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Caesar then the residue of F at C 0 can
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Symphony found by taking G the numerator
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function evaluated at Z 0 and dividing
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that by the denominator derivative at T
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0
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let's get back to the example that we
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looked at when we looked at a double
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tool Epic's equals one over Z minus one
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squared Z minus three it's a double pull
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at one but also it's simple pull at
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three this time we want to find the
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residue at three we can look at this
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function ask the function 1 over Z minus
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1 squared divided by the function Z
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minus 3 the function G which is 1 over Z
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minus 1 squared has an isolated
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singularity and one but is otherwise
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analytics in particular near 3 it's very
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much analytic for example if I look at a
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little disc around 3 and that disc the
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function G is entirely analytic the
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function H of Z which is U minus 3 is
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also analytic it's analytic everywhere
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it has a simple zero at 3 by the fact we
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can therefore find the residue of the
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function f which is G divided by H by
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simply plugging in 3 into G and plugging
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in 3 into the derivative of H well G of
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3 is 1 over 3 minus 1 squared 3 minus 1
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is negative 2 square that you get for it
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so this is 1/4 and H prime of 3 is equal
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to 1 because the derivative of H is
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simply equal to 1 everywhere
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therefore the residue of this function
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FXa 0 is 1/4 we could have just as well
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found the residue of F using the fact
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that 3 is a simple pool we could have
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applied that formula the calculation
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would have been quite similar this is
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just another way of finding residues in
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the next lecture we'll use the residue
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theorem to find some interesting
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integrals