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Consider the following integral:

$\displaystyle \int_{|z|=1} \frac{e^z}{(z+3)\sin(2z)}dz$.

To apply the Cauchy integral formula, I rewrite it as:

$\displaystyle \int_{|z|=1} \frac{ze^z}{z(z+3)\sin(2z)}dz$ and take

$\displaystyle f(z)=\frac{ze^z}{(z+3)\sin(2z)}$.

The problem now is that I would compute $f(0)$ in the next step, but $\sin(2\cdot 0)=0$, so the denominator of $f$ is undefined. How would I deal with this?

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up vote 3 down vote accepted

Hint: What type of singularity is $z = 0$ for the function $$f(z) = \frac{ze^z}{(z+3)\sin(2z)} = \frac{e^z}{z+3}\frac{z}{\sin(2z)}?$$

In other words: is there any reasonable way of defining $f(0)$, perhaps by some sort of limit?

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It's a removable singularity. So we could look at $\lim_{z\rightarrow 0} f(z) = \frac{1}{6}$. Do we now set $f(0)=\frac{1}{6}$ and carry on with the Cauchy integral formula? What would have happened if the singularity wasn't removable or if we had multiple singularities? – Chris Jul 23 '12 at 19:29
Yes, you can now set $f(0) = \frac{1}{6}$ and carry on. If the singularity is a pole, then you can often use the more general Cauchy Integral Formula $$\frac{1}{2\pi i}\int_C \frac{f(z)}{(z-a)^{n+1}}\,dz = \frac{f^{(n)}(a)}{n!}.$$ (I assume you've only been using the $n = 0$ version.) If the singularity is essential... I'm not sure if I can think of a general method. Residues sometimes help. – Jesse Madnick Jul 23 '12 at 19:38
For example, if you wanted to integrate $g(z) = \frac{1}{z^4 \sin z}$ around $|z| = 1$, you can write $g(z) = \frac{z/\sin z}{z^5}.$ Now the numerator has a removable singularity, and you can apply the Cauchy Integral Formula with $n = 4$. – Jesse Madnick Jul 23 '12 at 19:42

The following applies for some neighbourhood of zero, which is what we're interested in (the only possible singularity of the function within the unit circle) $$\frac{1}{3\left(1+\frac{z}{3}\right)}\frac{1}{\sin 2z}e^z=\frac{1}{3}\left(1-\frac{z}{3}+\frac{z^2}{9}-...\right)\frac{1}{2z-\frac{8z^3}{3!}+...}\left(1+z+\frac{z^2}{2!}+...\right)=$$ $$=\frac{1}{3}\left(1-z+\frac{z^2}{9}-...\right)\,\frac{1}{2z}\frac{1}{1-\frac{4z^2}{3}+...}\left(1+z+\frac{z^2}{2!}+...\right)=$$ $$=\frac{1}{6z}\left(1-z+\frac{z^2}{9}+...\right)\left(1+z+\frac{z^2}{2}+...\right)\left(1+\frac{4z^2}{3}+...\right)=\frac{1}{6z}+...$$ Thus, the function's residue at $\,z=0\,$ is $\,1/6\,$ and by the residue's theorem the integral's value is $$\frac{1}{6}2\pi i=\frac{\pi i}{3} $$

Of course, the residue of this pole (because that is what $\,z=0\,$ is: a simple pole, as can be easily checked) is waaaaaay easier to evaluate by the well-known formula $$Res_{z=0}(f)=\lim_{z\to 0}\frac{ze^z}{(z+3)\sin 2z}=\frac{1}{6}$$ but using power series can be serious fun...and pretty helpful sometimes if the pole's order is high.

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