# Computing a very messy contour integral

I'm hoping that someone might be able to help me with the following problem. I'll walk through my current work and indicate where I'm stuck.

Compute the contour integral:

$$\oint_{|z-1-i| = 5/4} \frac{Log(z)}{(z-1)^2} \, dz.$$

Clearly, the curve represents a circle of radius $5/4$ centered at the point $z_0 = 1 + i$. After drawing the curve, I've noticed that its interior contains two singularities, namely $z=0$ (where $Log(z)$ is undefined) and $z=1$ (where the denominator of the integrand is undefined).

Let $C$ be the curve represented by $|z-1-i| = 5/4$ and let $D$ be the interior of $C$. My first step is to split the region $D$ into two pieces:

$$D_1 = \left\{ z \in D \ | \ Im(z) < 1/2 \right\}$$

$$D_2 = \left\{ z \in D \ | \ Im(z) > 1/2 \right\}.$$

Let $C_1$ be the boundary of $D_1$ and let $C_2$ be the boundary of $D_2$. Now we have

$$\oint_{|z-1-i| = 5/4} \frac{Log(z)}{(z-1)^2} \, dz = \oint_{C_1} \frac{Log(z)}{(z-1)^2} \, dz + \oint_{C_2} \frac{Log(z)}{(z-1)^2} \, dz.$$

To compute the second integral, note that $Log(z)$ is analytic in $D_2$, and so we can apply the modified version of the Cauchy integral formula. We have

$$\oint_{C_2} \frac{Log(z)}{(z-1)^2} \, dz = 2\pi i \cdot f'(1)$$

where $f(z) = Log(z)$. Since $f'(z) = 1/z$, $f'(1) = 1$ and so the integral is equal to $2\pi i$.

Finally we need to compute the first integral. Note that $D_1$ contains a singularity of $Log(z)$, namely $z=0$. We construct a circle of radius $0.25$ centered at the origin. Since $Log(z)$ is analytic on the rest of $D_1$, the value of the whole integral is just the value of

$$\oint_{C'} \frac{Log(z)}{(z-1)^2} \, dz$$

where $C'$ is the circle of radius $0.25$ centers at the origin.

Here's where I'm running into trouble -- namely in parametrizing the circle $C'$. Am I even on the right path toward a solution?

• $z=0$ is not inside the contour. Commented Jul 24, 2014 at 10:48
• Ah! Silly mistake on my part. Sorry to waste your time. Commented Jul 24, 2014 at 11:09

Hint: Consider $w=z-1$. The integral becomes $$\oint_{|w-i|=5/4}\frac{\log(1+w)}{w^2}\mathrm{d}w$$ Recall the power series for $\log(1+w)=w-\frac{w^2}{2}+\frac{w^3}{3}-\dots$
Since $w=0$ is inside the contour, we can apply the Residue Theorem.
Comment: There is nothing wrong with your approach using Cauchy's Integral Formula. We are both actually looking at the first derivative of $\log(z)$ at $z=1$. However, you don't need to break up the contour.
Caveat: be careful about considering $\log(z)$ on contours which circle the origin.
$\hspace{4.5cm}$