- Laurent series of \(f\left ( z\right ) \) around point \(z_{0}\) is \(\sum _{n=-\infty }^{\infty }a_{n}\left ( z-z_{0}\right ) ^{n}\) and \(a_{n}=\frac {1}{2\pi i}{\displaystyle \oint } \frac {f\left ( z\right ) }{\left ( z-z_{0}\right ) ^{n+1}}dz\). Integration is around path enclosing \(z_{0}\) in counter clockwise.
- Power series of \(f\left ( z\right ) \) around \(z_{0}\) is \(\sum _{0}^{\infty }a_{n}\left ( z-z_{0}\right ) ^{n}\) where \(a_{n}=\frac {1}{n!}\left . f^{\left ( n\right ) }\left ( z\right ) \right \vert _{z=z_{0}}\)
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Problem asks to use Cauchy integral formula \({\displaystyle \oint \limits _{C}} \frac {f\left ( z\right ) }{z-z_{0}}dz=2\pi if\left ( z_{0}\right ) \) to evaluate another integral \({\displaystyle \oint \limits _{C}} g\left ( z\right ) dz\). Both over same \(C\). The idea is to rewrite \(g\left ( z\right ) \) as \(\frac {f\left ( z\right ) }{z-z_{0}}\) by factoring out the poles of \(g\left ( z\right ) \) that are outside \(C\) leaving one inside \(C\). Then we can write
\begin{align*}{\displaystyle \oint \limits _{C}} g\left ( z\right ) dz & ={\displaystyle \oint \limits _{C}} \frac {f\left ( z\right ) }{z-z_{0}}dz\\ & =2\pi if\left ( z_{0}\right ) \end{align*}
For example, to solve \({\displaystyle \oint \limits _{C}} \frac {1}{\left ( z+1\right ) \left ( z+2\right ) }dz\) around \(C\) unit circle. Rewriting this as \({\displaystyle \oint \limits _{C}} \frac {\frac {1}{z+2}}{\left ( z-\left ( -1\right ) \right ) }dz\) where now \(f\left ( z\right ) =\frac {1}{z+2}\) and now we can use Cauchy integral formula. So all what we have to do is just evaluate \(\frac {1}{z+2}\) at \(z=-1\), which gives \({\displaystyle \oint \limits _{C}} \frac {1}{\left ( z+1\right ) \left ( z+2\right ) }dz=2\pi i\). This works if \(g\left ( z\right ) \) can be factored into \(\frac {f\left ( z\right ) }{z-z_{0}}\) where \(f\left ( z\right ) \) is analytic on and inside \(C\). This would not work if \(g\left ( z\right ) \) has more than one pole inside \(C\).
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Problem asks to find \({\displaystyle \oint \limits _{C}} f\left ( z\right ) dz\) where \(C\) is some closed contour. For this, if \(f\left ( z\right ) \) had number of isolated singularities inside \(C\), then just use
\[{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=2\pi i\sum \text {residues of}\ f\left ( z\right ) \ \text {at each singularity inside }C \]
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Problem asks to find \(\int _{C}f\left ( z\right ) dz\) where \(C\) is some open path, i.e. not closed (if it is closed, try Cauchy), such as a straight line or a half circle arc. For these problem, use parameterization. This converts the integral to line integration. If \(C\) is straight line, use standard \(t\) parameterization, which is found by using
\begin{align*} x\left ( t\right ) & =\left ( 1-t\right ) x_{0}+tx_{1}\\ y\left ( t\right ) & =\left ( 1-t\right ) y_{0}+ty_{1}\end{align*}
where \(\left ( x_{0},y_{0}\right ) \) in the line initial point and \(\left ( x_{1},y_{1}\right ) \) is the line end point. This works for straight lines. Now use the above and rewrite \(z=x+iy\) as \(z\left ( t\right ) =x\left ( t\right ) +iy\left ( t\right ) \) and then plug-in in this \(z\left ( t\right ) \) in \(f\left ( z\right ) \) to obtain \(f\left ( t\right ) \), then the integral becomes
\[ \int _{C}f\left ( z\right ) dz=\int _{t=0}^{t=1}f\left ( t\right ) z^{\prime }\left ( t\right ) dt \]
And now evaluate this integral using normal integration rules. If the path is a circular arc, then no need to use \(t\), just use \(\theta \). Rewrite \(x=re^{i\theta }\) and use \(\theta \) instead of \(t\) and follow same steps as
above.
- Problem gives \(u\left ( x,y\right ) \) and asks to find \(v\left ( x,y\right ) \) in order for \(f\left ( x,y\right ) =u\left ( x,y\right ) +iv\left ( x,y\right ) \) to be analytic in some region. To solve these, use Cauchy Riemann equations. Need to use both equations. One equation will introduce a constant of integration (a function) and the second equation is used to solve for it. This gives \(v\left ( x,y\right ) \). See problem 2, HW 2, Physics 501 as example.
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Problem asks to evaluate \({\displaystyle \oint \limits _{C}} \frac {f\left ( z\right ) }{\left ( z-z_{0}\right ) ^{n}}dz\) where \(n\) is some number. This is the order of the pole, and \(f\left ( z\right ) \) is analytic on and inside \(C\). Then use the Cauchy integral formula for higher pole order. \({\displaystyle \oint \limits _{C}} \frac {f\left ( z\right ) }{\left ( z-z_{0}\right ) ^{n}}dz=2\pi i\ \operatorname {Residue}\left ( z_{0}\right ) \). The only difference here is that this is pole of order \(n\). So to find residue, use
\begin{align*} \operatorname {Residue}\left ( z_{0}\right ) & =\lim _{z\rightarrow z_{0}}\frac {d^{n-1}}{dz^{n}}\frac {\left ( z-z_{0}\right ) ^{n}}{\left ( n-1\right ) !}\frac {f\left ( z\right ) }{\left ( z-z_{0}\right ) ^{n}}\\ & =\lim _{z\rightarrow z_{0}}\frac {d^{n-1}}{dz^{n}}\frac {f\left ( z\right ) }{\left ( n-1\right ) !}\end{align*}
- Problem gives \(f\left ( z\right ) \) and asks to find branch points and branch cuts. One way is to first find where \(f\left ( z\right ) =0\) and for each zero, make a small circle around it, starting from \(\theta =0\) to \(\theta =2\pi \). If the function at \(\theta =0\) has different value from \(\theta =2\pi \), then this is a branch point. Do this for other zeros. Then connect the branch points. This will give the branch cut. It is not always clear how to connect the branch point though, might need to try different
ways. For example \(f\left ( z\right ) =\sqrt {z^{2}+1}\) has two zeros at \(z=\pm i\). Both turn out to be branch points. The branch cut is the line between \(-i\) to \(+i\) on the imaginary axis.
- Problem gives a series \(\sum _{n=0}^{\infty }a_{n}z^{n}\) and asks to find radius of convergence \(R\). Two ways, find \(L=\lim _{n\rightarrow \infty }\frac {\left \vert a_{n+1}\right \vert }{\left \vert a_{n}\right \vert }\) and then \(R=\frac {1}{L}\). Another way is to find \(L\) using \(L=\lim _{n\rightarrow \infty }\left \vert a_{n}\right \vert ^{\frac {1}{n}}\).
- Problem gives integral \(\int _{0}^{2\pi }f\left ( \theta \right ) d\theta \) and asks to evaluate using residues. We start by converting everything to \(z\) using \(z=e^{i\theta }\) using \(\left \vert z\right \vert =1\). No need to use \(z=re^{i\theta }\). The idea is to convert it to \({\displaystyle \oint } f\left ( z\right ) dz\) which then we can use \({\displaystyle \oint } f\left ( z\right ) dz=2\pi i\sum \) residues inside. Replace \(f\left ( \theta \right ) \) to become \(f\left ( z\right ) \), this could require using Euler relation such as \(\cos n\theta =\frac {z^{n}+z^{-n}}{2}\) and similar for \(\sin \theta \). Now all what is needed is to find residues of any poles inside the unit circle. Do not worry about poles outside the unit circle. To find
residues use short cut tricks. No need to find Laurent series.
For an example, to evaluate \(\int _{0}^{2\pi }\frac {1}{5+4\cos \theta }d\theta \), then \(\frac {1}{5+4\cos \theta }\) becomes \(\frac {1}{\left ( 2z+1\right ) \left ( z+2\right ) }\) and there is only one pole inside unit circle, at \(z=-\frac {1}{2}\).
- Problem gives integral \(\int _{0}^{\infty }f\left ( \theta \right ) d\theta \) and asks to evaluate using residues. The contour here goes from \(-R\) to \(+R\) and then a semi circle in upper half plane. This works for even \(f\left ( \theta \right ) \) since we can write \(\int _{0}^{\infty }f\left ( \theta \right ) d\theta =\frac {1}{2}\int _{-\infty }^{\infty }f\left ( \theta \right ) d\theta \). If there is a pole inside the upper half plane, then the integral over the semi circle is \(2\pi i\) times the sum of residues. If there is a pole on the real line, then make a small semi circle around pole, say at \(z=a\) and then the integral for the small semi
circle is \(-\pi i\) times the residue at \(a\). The minus sign here is due to moving clock wise on the small circle.
- Problem gives a series \(\sum _{n=0}^{\infty }a_{n}z^{n}\) and asks if it is uniformly convergent. For general series, use the M-test. But for this kind of series, just find radius of convergence as above using ratio test, and if it is absolutely convergent, then say it converges uniformly for \(\left \vert z\right \vert \leq r<R\). It is important to write it this way, and not just \(\left \vert z\right \vert <R\).
- Problems gives \(\sum _{n=0}^{\infty }a_{n}\) and asks to find the sum. Sometimes this trick works for some series. For example the alternating series \(\sum _{n=1}^{\infty }\left ( -1\right ) ^{n+1}\frac {1}{n}=1-\frac {1}{2}+\frac {1}{3}-\frac {1}{4}+\cdots \), then write it as \(x-\frac {x^{2}}{2}+\frac {x^{3}}{3}-\frac {x^{4}}{4}+\cdots \) which is the same when \(x=1\), and now notice that this is the Taylor series for \(\ln \left ( 1+x\right ) \) which means when \(x=1\) then \(1-\frac {1}{2}+\frac {1}{3}-\frac {1}{4}+\cdots =\ln \left ( 2\right ) \).
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Problem gives \(f\left ( z\right ) \) and asks to find residue at some \(z=z_{0}\). Of course we can always expand \(f\left ( z\right ) \) around \(z=0\) using Laurent series and find the coefficient of \(\frac {1}{z}\). But this is too much work. Instead, if \(f\left ( z\right ) \) has a simple pole of order one, then we use
\[ R\left ( z_{0}\right ) =\lim _{z\rightarrow z_{0}}\left ( z-z_{0}\right ) f\left ( z\right ) \]
In general, if \(f\left ( z\right ) =\frac {g\left ( z\right ) }{h\left ( z\right ) }\) then there are two cases. If \(h\left ( z_{0}\right ) =0\) or not. If \(h\left ( z_{0}\right ) \neq 0\), then we can just use the above. For example, if \(f\left ( z\right ) =\frac {z}{\left ( 2z+1\right ) \left ( 5-z\right ) }\) and we want the residue at \(z_{0}=5\), then since it simple pole, then using
\begin{align*} R\left ( 5\right ) & =\lim _{z\rightarrow 5}\left ( z-5\right ) \frac {z}{\left ( 2z+1\right ) \left ( 5-z\right ) }\\ & =\lim _{z\rightarrow 5}\frac {-z}{\left ( 2z+1\right ) }\\ & =-\frac {3}{11}\end{align*}
But if \(h\left ( z_{0}\right ) =0\) then we need to apply La’Hopital like this. If \(f\left ( z\right ) =\frac {\sin z}{1-z^{4}}\) and we want to find residue at \(z=i\). Then do as above, but with extra step, like this
\begin{align*} R\left ( i\right ) & =\lim _{z\rightarrow i}\left ( z-i\right ) \frac {\sin z}{1-z^{4}}\\ & =\left ( \lim _{z\rightarrow i}\sin z\right ) \left ( \lim _{z\rightarrow i}\left ( z-i\right ) \frac {1}{1-z^{4}}\right ) \\ & =\sin i\left ( \lim _{z\rightarrow i}\frac {\left ( z-i\right ) }{1-z^{4}}\right ) \qquad \text {Now apply La'Hopital}\\ & =\sin i\left ( \lim _{z\rightarrow i}\frac {1}{-4z^{3}}\right ) \\ & =\frac {\sin i}{-4i^{3}}\\ & =\frac {1}{4}\sinh \left ( 1\right ) \end{align*}
Now if the pole is not a simple pole or order one,.say of order \(m\), then we first multiply \(f\left ( z\right ) \) by \(\left ( z-z_{0}\right ) ^{m}\) then differentiate the result \(m-1\) times, then divide by \(\left ( m-1\right ) !\), and then evaluate the result at \(z=z_{0}.\) in other words,
\[ R\left ( z_{0}\right ) =\lim _{z\rightarrow z_{0}}\frac {1}{\left ( m-1\right ) !}\frac {d^{m-1}}{dz^{m-1}}\left ( \left ( z-z_{0}\right ) ^{m}f\left ( z\right ) \right ) \]
For example, if \(f\left ( z\right ) =\frac {z\sin z}{\left ( z-\pi \right ) ^{3}}\) and we want residue at \(z=\pi \). Since order is \(m=3\), then \begin{align*} R\left ( z_{0}\right ) & =\lim _{z\rightarrow \pi }\frac {1}{2!}\frac {d^{2}}{dz^{2}}\left ( \left ( z-\pi \right ) ^{3}\frac {z\sin z}{\left ( z-\pi \right ) ^{3}}\right ) \\ & =\lim _{z\rightarrow \pi }\frac {1}{2}\frac {d^{2}}{dz^{2}}\left ( z\sin z\right ) \\ & =\lim _{z\rightarrow \pi }\frac {1}{2}\left ( -z\sin z+2\cos z\right ) \\ & =-1 \end{align*}
The above methods will work on most of the HW problems I’ve seen so far but If all else fails, try Laurent series, that always works.