Evaluate $ \int \frac {1} {z^2+1} dz $ along the contour $\Gamma$. (Gamma is some closed circle centered around i, no specified radius, and is oriented counter clockwise.)

So far, I've used factored the expression into $ \int \frac 1 {(z+i)(z-i)}$, then used partial fraction decomposition, and got $ \frac 1 {2i}[ \frac{1} {z+i}-\frac{1} {z-i}] $.

Now I can take the integral of both of these separate terms from 0 to 2pi, but I'm having a lot of trouble with parameterization of this curve before I integrate.

I assumed since it's a closed contour centered at i, I would parameterize it by $z(t)=i+e^{it}$? However, when I substitute this into the integral and multiply by it's derivative $z'(t) = ie^{it}$, I don't get the right answer. The book says I should be getting zero for the first integral, but I don't even know how to do this integral.

$$ \int_{\Gamma}\frac{1}{z+i}dz =\int_0^{2\pi}(\frac{1}{(i+e^{it})+i})(ie^{it})dt=\int_0^{2\pi}\frac {ie^{it}}{e^{it}+2i}dt$$

Now I have absolutely no clue how to do this integral, but supposedly it is supposed to come out to 0, and the second integral of $$\int_0^{2\pi}\frac{1}{z-i}$$ should be equal to $ 2\pi i$. They then total together using the 3rd equation from the top, and 2i cancels, and the total integral becomes $ \pi $. That should be the total answer.

My question mainly is, how do I do that first integral of $\frac{1}{}

  • 1
    $\begingroup$ $(e^{i t}+2i)'=i e^{i t}$ $\endgroup$
    – tired
    Oct 4, 2016 at 13:28
  • 2
    $\begingroup$ You have a sign error, $\frac{1}{z^2+1} = \frac{1}{2i}\bigl(\frac{1}{z-i} - \frac{1}{z+i}\bigr)$. You probably have heard a theorem that lets you evaluate the integral without parametrising $\Gamma$. $\endgroup$ Oct 4, 2016 at 13:29
  • $\begingroup$ I apologize, I meant to have transformed the 1/2i to i/2, therefore distributing the negative sign on them. Thank you for pointing that out. $\endgroup$ Oct 4, 2016 at 20:10

1 Answer 1


If one wishes to proceed with "brute force," then we can write

$$\begin{align} \oint_{|z-i|=1} \frac{1}{z+i}\,dz&=\int_0^{2\pi}\frac{ie^{it}}{e^{it}+i2}\,dt\\\\ &=\int_0^{2\pi}\frac{\cos(t)+i(1+2\sin(t))}{5+4\sin(t)}\,dt\\\\ &=\color{blue}{\int_0^{2\pi}\frac{2\cos(t)}{5+4\sin(t)}\,dt}+\color{red}{i\int_{0}^{2\pi}\frac{1+2\sin(t)}{5+4\sin(t)}\,dt}\\\\ &=\color{blue}{\left.\left(\frac12\log(5+4\sin(t))\right)\right|_0^{2\pi}}+\color{red}{\left.i\arctan\left(\frac{2+\sin(t)}{\cos(t)}\right)\right|_{-\pi/2}^{\pi/2}+\left.i\arctan\left(\frac{2+\sin(t)}{\cos(t)}\right)\right|_{\pi/2}^{3\pi/2}}\\\\\ &=0 \end{align}$$


Since $\frac{1}{z+i}$ is analytic for $|z-i|\le 1$, then Cauchy's Integral Theorem gives immediately


as expected!

  • $\begingroup$ Thank you this is precisely what I was looking for. The issue was my book had a theorem along the lines of $ \int_{\Gamma} (z-z_0)^n = 0$ if $n \neq 1$ and it equals $2\pi i$ if $n=1$ and I was very confused because I figured $\frac {1} {z+i} $ would have an n = 1, and therefore be non-zero. $\endgroup$ Oct 4, 2016 at 20:15
  • $\begingroup$ You're welcome! My pleasure Ryan. Pleased that this was helpful. Note that if the contour encloses the pole, then the integral is $2\pi i$. $\endgroup$
    – Mark Viola
    Oct 4, 2016 at 20:19

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