In the exercise questions(just exercise not homework) for a Complex analysis course we are asked the following;
Suppose we have $f$ analytic on the disc $D(0,2)$ show that;
$$
\int_C\frac{f(w)}{w-i/2}\text{d}w =
\int_{\Delta}\frac{f(w)}{w-i/2}\text{d}w.
$$
Where $C$ is the unit circle centered at zero, counter clockwise, and $\Delta$ is the triangle, counter clockwise, with corners $i,\pm 1-i$.
Here is what I have so far;
Since $f$ is analytic on $D(0,2)$, Cauchy's integral applies formula to $C$ and since and $i/2$ in the interior of $C$ we have
$$
f(i/2)=\frac{1}{2\pi i}\int_C\frac{f(w)}{w-i/2}\text{d}w.
$$
So it is sufficient to show that
$$
f(i/2)=\frac{1}{2\pi i}\int_{\Delta}\frac{f(w)}{w-i/2}\text{d}w.
$$
But now consider the difference quotient
$$
g(i/2) =
\begin{cases}
\frac{f(w)-f(i/2)}{w-i/2},& w\neq i/2\\
f'(i/2)&w=i/2
\end{cases}
$$
Now one can show that $g$ is analytic on the disc $D(0,2)$ and so by the closed curve theorem we know
$$
\int_{\Delta}\frac{f(w)-f(i/2)}{w-i/2}\text{d}w = 0
$$
which gives
$$
\int_{\Delta}\frac{f(w)}{w-i/2}\text{d}w = f(i/2)\int_{\Delta}\frac{1}{w-i/2}\text{d}w
$$
So it is sufficient to show that
$$
\int_{\Delta}\frac{1}{w-i/2}\text{d}w = 2\pi i
$$
Now since $|i/2|<w$, for $w$ on $\Delta$ we have that $\frac{|i/2|}{|w|}<1$
$$
\frac{1}{w-i/2} = \frac{1}{w}\frac{1}{1-\frac{i}{2w}}=\frac{1}{w}\sum_{k=0}^{\infty}\bigg(\frac{i}{2w}\bigg)^k
$$
and the power series converges uniformly so we can write
$$
\int_{\Delta}\frac{1}{w-i/2}\text{d}w =\frac{1}{w}\sum_{k=0}^{\infty}\int_{\Delta}\bigg(\frac{i}{2w}\bigg)^k.
$$
But now for each $k\geq 1$ we have that $\frac{-1}{kw^k}$ is the analytic anti-derivative of $\frac{1}{k^{k+1}}$ and so by the closed curve theorem we have
$$
\int_{\Delta}\frac{1}{w-i/2}\text{d}w = \int_{\Delta}\frac{1}{w}dw.
$$
AAAAANDD… now i'm stuck. Im not sure how to show that this integral is equal to $2\pi i$… I have a feeling we need the complex logarithm, but we haven't learnt that yet.
Have I made any mistakes in the working above? Is there a way forward that does not need the complex logarith? Do we even need the complex logarithm?
Cheers in advance for any help.
Best Answer
Here is an easier approach that avoids computing the integrals and using Cauchy's integral formula. You can draw a picture of $C$ and $\Delta$ and then notice that $\int_C-\int_\Delta$ is the same as the sum over four integrals over certain closed contours. Each of these contours lies in a half-disk where the function $z\mapsto \frac{f(z)}{z-i/2}$ is analytic, so integrals over those contours are zero, and so the original integrals coincide.