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?

  • 4
    $\begingroup$ $z=0$ is not inside the contour. $\endgroup$
    – pshmath0
    Commented Jul 24, 2014 at 10:48
  • $\begingroup$ Ah! Silly mistake on my part. Sorry to waste your time. $\endgroup$
    – Ryan Peden
    Commented Jul 24, 2014 at 11:09

1 Answer 1


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.

$\hspace{4.5cm}$enter image description here


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