33 - The Residue Theorem
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welcome to the third lecture in the
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seventh week of our course analysis of a
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complex kind today we'll learn about a
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really powerful theorem and leave the
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rezident theorem let's start with the
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motivation then we remind you about
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isolated singularities remember that if
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F has an isolated singularity at a point
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say zero that means F is analytic in
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some disk centered it to zero with the
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exception of C 0 then F is a Laurent
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series expansion in that punctured disk
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and the Laurent series expansion looks
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like this
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we have coefficients a K and power Z
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minus Z 0 to the K for K from negative
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infinity to infinity and the series
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converges locally uniformly in this
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punctured disk and it converges
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uniformly as soon as we stay away from
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the boundary of that disk now notice the
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following what happens when we integrate
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F over a circle of radius Rho centered
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at Z 0 where Rho is a little bit less
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than R so this red curve that is drew
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it's a compact set inside the punctured
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disk and therefore the convergence of
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the series that represents F is uniform
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on that curve that means when we're
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integrating F over this curve we're
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really integrating this sum over the
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curve because the convergences uniform
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we can interchange the summation and the
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integration that is typically not the
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case but under uniform convergence
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you're allowed to interchange and
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therefore we find ourselves with a sum
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on the outside of the integration on the
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inside that makes it quite easy to
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evaluate the integral over f DZ because
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now all we have to do is find all these
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individual integrals that have nothing
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to do with F itself we simply have to
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find out what does the integral over as
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you - easier to the K over the circle of
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radius Rho centered at 0 we distinguish
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two cases namely the case where K is not
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equal to negative 1 and the case where K
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is equal to negative 1 if K is not equal
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to negative 1 then this function Z minus
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2 0 to the K actually has an
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antiderivative that is analytic in the
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entire complex plane with the except
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zero and the antiderivative is one over
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K plus 1 C minus C 0 to the k plus 1
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because when they differentiate this
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function on the right the k plus 1 comes
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down can't sever the 1 over k plus 1 and
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the exponent becomes a gain which is
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exactly what I want the only time where
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that doesn't works when K is equal to
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negative 1 because I would be divided by
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0 in that case so we have an
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antiderivative which means the integral
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because we're integrating over a closed
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curve amounts to being 0 when K is equal
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to negative 1 however I don't have an
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antiderivative of that type and
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therefore the integral needs to be
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evaluated in a different way we can
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certainly use the cosy integral formula
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which is complete overkill but we can
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also calculate the integral directly let
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me remind you how to find such integrals
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what we need is we need to parameterize
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the curve over which we want to
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integrate we're integrating over a
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circle of radius row centered at a point
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Z 0 one way to memorize such a circle is
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with a parameterization gamma of T
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equals C 0 plus Rho e to the I T we'll
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need the derivative of this
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parameterization that it is game I prime
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of T equals Rho I die T now we simply
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plug that in so instead of Z I read Z 0
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plus Rho e to the I T and subtract this
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is 0 negative multiplied with a
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derivative of the parameters a and
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notice that a lot of things cancel out
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in the denominator of the Z 0 goes away
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and I'm left the throat e to the I T and
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now even the row if that I T cancels out
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so all I'm left with is the integral
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from 0 to 2 pi of I DT and that equals 2
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pi so when K is equal to a negative 1
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the integral that I'm interested in
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equals 2 pi I now remember what we
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actually were trying to find we're
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trying to find the integral over f of ZD
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c where f had
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isolated singularity at z0f was analytic
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and that punctured discs punctured at su
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0 of some radius R with an isolated
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singularity at C theorem we have looked
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at this Laurent series expansion and we
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were able to bring the summation to the
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outside and the integration to the
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inside and we're now left with this
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integral u minus a 0 equals Rho over as
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u minus e0 to the kdz we saw that unless
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K is equal to negative 1 this integral
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is not even there so this whole integral
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we just realize this equal to 0 as long
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as K is not equal to negative 1 and so
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there's only one term left in the sum
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namely that term where K is equal to
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negative 1 when K is equal to negative 1
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the integral equals to PI and the only
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thing that's then left is the
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coefficient here namely the a minus 1
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coefficient in that case and so we end
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up with 2 pi I times a negative 1 in
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other words if I integrate a function
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with an isolated singularity around that
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singularity for the curve that's small
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enough to fit into the disk where the
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function is otherwise an analytic then I
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get 2 pi I times a minus 1 where a minus
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1 is the coefficient of the 1 over Z
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minus is 0 term and the overall series
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expansion so therefore this a minus 1
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gets very special attention and that's
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why I've been having you calculate a
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minus 1 and so many examples in the
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previous in-video quizzes this a minus 1
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is so special that it even gets any if f
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has an isolated singularity at Z 0 with
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this Laurent series expansion right here
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then the residue of F at C 0 is exactly
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this a minus 1 term so this a minus 1
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term is called the residue that's user
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let's look at some examples we already
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calculated the Laurent series expansion
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of the function 1 over Z minus 1 times Z
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minus 2 we know this function has
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isolated singularities at Z equals 1 and
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Z equals 2 and so there's lots of
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different ways to find any way in which
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the function is indeed analytic in which
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we can find an overall series expansion
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if we were interested in the residue of
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this function at 0.1 then we defined the
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Laurent series expansion in a poacher
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disc centered at 1/2 for example this
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one right here let the disc be used and
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we found that in that disc F has a
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Laurent series expansion which is
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negative 1 over Z minus 1 times the sum
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that you see right here the sum is only
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positive powers of Z minus 1 the only
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negative power of seeing when this one
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is right here and since it's the powers
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u minus 1 to the negative 1 we see the
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coefficient a negative 1 right up here
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in other words a minus 1 is equal to
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negative 1 times the residue of F and 1
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is a negative 1 if on the other hand we
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wanted to find the residue of this
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function at the isolated singularity at
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2 we need to find a liberal series
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expansion and it punctured is centered
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it to again we do a partial fraction
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decomposition can we find that the
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Laurent series is 1 over Z minus 2 minus
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this sum right here which contains only
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positive powers of Z minus 2 and
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therefore this term here represents the
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term for K equals negative 1 and
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therefore the one right here that's my a
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negative 1 for this velocity so the
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residue of this function f at the point
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2 is 1 so again here's the procedure if
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you want to find the residue of a
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function in isolated singularity what
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you need to do is you need to find a
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Laurent series expansion centered at
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that singular point and then read off
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the a negative 1 term here's another
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example the function f of Z equals sine
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Z over Z to the 4th
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we sound that this function has a pole X
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equal zero so it's an isolated
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singularity and we found it's LeBron
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series to be 1 over 2 cubed plus minus 1
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over 3 factorial times 1 over Z and so
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forth so here we have two terms with
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negative powers of Z but this term in
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particular is the K equals negative 1
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term and this is the coefficient a minus
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1 right there because it's the
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coefficient in front of the 1 over C
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term and therefore the residue of F at 0
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is minus 1 over 3 factorial which is
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minus 1/6
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here's your another example the function
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cosine of 1 over C we saw this function
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as an essential singularity at 0 so the
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overall series expansion has infinitely
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many terms with negative powers of Z but
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we notice these are all even powers of Z
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there is no 1 over Z term therefore the
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coefficient of the 1 over Z term must be
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0 and therefore the residue of this
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function at 0 is equal to 0 how about
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sine of 1 over T that function also has
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an essential singularity at the origin
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but the Laurent series expansion has all
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odd powers of Z and indeed there is a 1
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over Z term in the Laurent series
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expansion and therefore the residue of
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that at 0 is equal to 1 now remember the
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function cosine of Z minus 1 over Z
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squared we notice that that function has
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a removable singularity at the origin
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because when you look at the Laurent
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series expansion there's no negative
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powers of Z left therefore again there
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is no 1 over z term in the residue of F
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of 0 equals 0 and here's a final example
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the function 1 over Z squared plus 1 the
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denominator factors into Z minus I times
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Z plus I and again we can do a partial
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fraction decomposition I'm going to do
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it the similar way that I showed you a
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few lectures ago I'm going to write the
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numerator is Z plus I minus Z minus I
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the Z's cancel out with each other and I
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minus minus I is 2 I so my numerator is
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now 2
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instead of one but if I divide by 2 I
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the new two on the numerator cancels out
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with the denominator of 2 I the reason I
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did this is so that I can cancel things
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out again I can now pull the fraction
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apart at the negative sign and I notice
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that when I look at the first fraction
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that Z plus I term cancels out and I'm
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left with 1 over Z minus 5 and if I look
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at the second fraction does hue - I
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terms cancel out and I'm left with 1
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over Z plus I 1 4 2 is in front of it
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and so here's my partial fraction
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decomposition of the function f this
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function has isolated singularities at Z
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equals I and Z equals negative 5 we want
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to find the residue at each of those
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points so near I I need to write this
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function in terms of powers of Z minus I
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I already have 1 over 2i times 1 over Z
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minus I the rest represents something
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that is analytic near I I don't even
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have to bother finding the exact
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representation and no one analytic
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function has a Taylor series expansion
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and so I can write this analytic
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function as a Taylor series K from 0 to
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infinity a K Z minus I to the K and
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together these two must form the Laurent
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series by uniqueness above the wrong
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series therefore the term for K equals
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negative 1 is the one here up front and
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a minus 1 is 1 over 2i which equals
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minus 1/2 I similarly when I look at F
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via negative I my goal is to write f as
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a Laurent series with powers of Z plus I
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I already have one such power of C plus
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I right here together with this negative
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sign of course into 1 over 2i in front
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of it so that you can see right here and
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the rest
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it's an analytic function near negative
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I which again will have a power series
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representation a K Z plus I to the K for
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K from 0 to infinity together the series
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and the first term form the Laurent
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series expansion of 1 over Z squared
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plus 1 near negative I and therefore
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this must be might a negative 1 term for
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this particular Laurent series
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therefore the residue of F at negative I
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is minus 1 over 2i which is 1/2
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here finally is the resident there the
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powerful theorem that this lecture is
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all about so both D is simply connected
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domain and to those F is an analytic
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function in this simply connected domain
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except for isolated singularities so
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here's my domain D and I the note of the
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isolated singularities with these little
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X that CD is simply closed curve and E
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and suppose C 1 through Zn of those
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isolated singularities of F that lie
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inside of this curve should not pass
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through any of the singularities -
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here's my curve C and it surrounds these
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two isolated singularities that are
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denoted C 1 and C 2 then theorem says
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the integral of F over this curve C is
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equal to 2 pi times the sum of the
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residues of F at the point CK that are
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inside the curve C in the particular
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example I drew here it would be simply
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getting 2 pi I times the residue of F at
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C 1 plus the residue of F at C 2 we've
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seen something like this already we
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notice that if we integrate a function
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with an isolated singularity around the
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circle centered at the isolated
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singularity and inside the region when
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function is analytic then what we get as
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the integral is actually exactly 2 pi I
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times the residue of function at the
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isolated singularity so the residue
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theorem is an extension of this fact
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that we just discovered
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here's an example let's again look at
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the function 1 over Z squared plus 1
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which is analytic in the entire complex
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plane with the exception of the isolated
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singularities at I and negative Phi I do
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a bunch of curves c1 c2 c3 and c4 in
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here
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suppose we first integrate over the
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curve C 1 just the red curve by the
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residue theorem I need to find 2 pi I
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times the residue of the function at the
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point that's surrounded by the curve
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which in this case is just a point I on
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the point that's surrounded by the curve
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C 1 and we already found that residue
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it's minus 1/2 I so altogether 2 is
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canceled out the eyes multiply to
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negative 1 cans log the other negative
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ones of the residue is equal to PI if I
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use the curve C 2 instead then this
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curve C 2 surrounds the point negative I
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and therefore I need to look at 2 pi I
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times the residue of F n negative I
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which we also found it's 1/2 PI and I
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find the integral equals negative pi now
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if I integrate it over the curve C 3 I
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noticed that her surrounds 2 isolated
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singularities namely I and negative I so
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if I integrate around C 3 I need to look
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at 2 pi I times the sum of those two
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residues but the two residues are minus
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1/2 and 1/2 I and they cancel each other
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out together I get that the integral
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over C 3 f of zdz equals 0 finally let's
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look at this blue curve right here it
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doesn't surround any singularities and
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so the sum is empty and therefore I get
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the integral over C 4 is equal to 0 here
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again is this very powerful theorem that
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we just learned about in a simply
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connected domain in which f is analytic
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except for isolated singularities with a
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simple closed curve or on which we are
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integrating we can find the integral
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over f over this curve as 2 pi times the
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sum of the residues of those
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singularities that are surrounded by sea
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in order to be able to really take
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advantage of this very powerful theorem
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we'll need some more strategies and
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techniques that will help
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to calculate residues so far we've been
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calculating entirely wrong series in
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order to find the residue of a function
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at a singularity but there are actually
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other ways to find residues and that's
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what we'll learn about in the next
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lecture