Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Calculate $\int_{\gamma}\frac{\Re(z)}{z-\frac{1}{2}}dz$ and $\int_{\gamma}\frac{\Im(z)}{z-\frac{1}{2}}dz$ when $\gamma$: $|z|=1$ is positively oriented.

This is what I have tried to do, starting with the first line integral.

Since $\frac{\Re(z)}{z-\frac{1}{2}}$ is not analytical/holomorphic at any point in the plane - it does not satisfy the Cauchy-Riemann equations - we cannot use Cauchy's integral formulae immediately. However, $|z|=1$ implies that the conjugate to $z$, $z*=\frac{1}{z}$, because $1=|z|^{2}=zz*$. Thus, we may write: \begin{align*} \int_{\gamma}\frac{\Re(z)}{z-\frac{1}{2}}dz &=\frac{1}{2}\int_{\gamma}\frac{z+\frac{1}{z}}{z-\frac{1}{2}}dz \\ &=\frac{1}{2}\int_{\gamma}\frac{z^{2}+1}{z(z-\frac{1}{2})}dz \\ &=\frac{1}{2}\int_{\gamma}(1-\frac{\frac{z}{2}+1}{z(z-\frac{1}{2})}dz \\ &=\frac{1}{2}\int_{\gamma}dz-\frac{1}{2}\int_{\gamma}\frac{\frac{z}{2}+1}{z(z-\frac{1}{2})}dz \\ &=-\frac{1}{2}\int_{\gamma}\frac{-2}{z}+\frac{\frac{5z}{2}}{z-\frac{1}{2}}dz \\ &= \int_{\gamma}\frac{1}{z}dz-\frac{5}{4}\int_{\gamma}\frac{z}{z-\frac{1}{2}}dz \\ &=2\pi{i}(1-\frac{5}{8}) \\ &=\frac{3\pi{i}}{4} \end{align*}

Note that I rewrote the right hand side two times, using partial fractions. Similarly,I have attempted to the second line integral, however, this time a only used partial fractions once.

\begin{align*} \int_{\gamma}\frac{\Im(z)}{z-\frac{1}{2}}dz &= \frac{1}{2i}\int_{\gamma}\frac{z-\frac{1}{z}}{z-\frac{1}{2}}dz \\ &=\frac{1}{2i}\int_{\gamma}\frac{z^{2}-1}{z(z-\frac{1}{2})}dz \\ &=\frac{1}{2i}\int_{\gamma}(1-\frac{2}{z}+\frac{3}{z-\frac{1}{2}})dz \\ &=\frac{1}{2i}\int_{\gamma}dz-\frac{1}{2i}\int_{\gamma}\frac{-2}{z}dz-\frac{1}{2i}\int_{\gamma}\frac{3}{z-\frac{1}{2}}dz \\ &= \frac{1}{i}\int_{\gamma}\frac{1}{z}dz-\frac{3}{4i}\int_{\gamma}\frac{1}{z-\frac{1}{2}}dz \\ &=2\pi{i}(\frac{1}{i}-\frac{3}{4i}) \\ &=\frac{5\pi}{4}\end{align*}

As I have not studied complex analysis, I am unsure whether any of these two answers are correct. I would very much appreciate if someone more experienced person could help me check, most likely correct, my calculations.

share|cite|improve this question

You have some problem on partial fraction $\displaystyle \frac{z^2 + 1}{z(z-1/2)} \neq 1 - \frac{\frac z 2 + 1}{z(z-1/2)}$.

This should be $\displaystyle \frac{z^2 + 1}{z(z-1/2)} = 1 + \frac{\frac z 2 + 1}{z(z-1/2)} = 1+\frac{5/2}{z - 1/2} - \frac 2 z$.
And this should give you $\displaystyle \frac{2 \pi i}{2} ( \frac 52 - 2) = \pi/2$

On the other, $\displaystyle \frac{z^2 - 1}{z(z-1/2)} = 1 + \frac 2 z - \frac{3/2}{z - 1/2}$ while you have written $\displaystyle \frac{z^2 - 1}{z(z-1/2)} = (1-\frac{2}{z}+\frac{3}{z-\frac{1}{2}})dz $

Also that on here $$=\frac{1}{2i}\int_{\gamma}(1-\frac{2}{z}+\frac{3}{z-\frac{1}{2}})dz \\ =\frac{1}{2i}\int_{\gamma}dz-\underbrace{\frac{1}{2i}\int_{\gamma}\frac{-2}{z}dz-\frac{1}{2i}\int_{\gamma}\frac{3}{z-\frac{1}{2}}dz}_{\text{why - here ??}} \\ $$ This is wrong step.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.