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I have to evaluate:

$$I=\int_{\gamma} \frac{\Re(z)}{2z-i} dz$$

where $\gamma$ is the unit circle centered at $0$. Cauchy's integral formula tells us that:

$$f^{(n)}(a) = \frac{n!}{2 \pi i} \int_{\gamma} \frac{f(z)}{(z-a)^{n+1}} dz$$

But this only works for $f$ holomorphic, and I'm finding it tough to work out how to apply it to this problem, since $\Re$ isn't a holomorphic function anywhere, let alone in the disc enclosed by $\gamma$.

Is there some sort of trick to pull out some nice factors so I can get the integral into a form I can work with?

pizzaroll
  • 993
  • On the unit circle, can you express $\overline{z}$ in a way that gives you a holomorphic function on some neighbourhood of the circle? – Daniel Fischer Nov 11 '16 at 16:42

1 Answers1

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Note that $\text{Re}(z)=\frac12(z+\bar z)$. Furthermore, note that for $z\in \gamma$, $\bar z=\frac1z$. Then, we can write

$$\begin{align} \oint_\gamma \frac{\text{Re}(z)}{2z-i}\,dz&=\frac12 \oint_\gamma \frac{z}{2z-i}\,dz+\frac12 \oint_\gamma \frac{\bar z}{2z-i}\,dz\\\\ &=\frac12\oint_\gamma \frac{z}{2z-i}\,dz+\frac12 \oint_\gamma \frac{1}{(2z-i)z}\,dz\\\\ &=\frac14\oint_\gamma \frac{z}{z-i/2}\,dz+\frac i2 \oint_\gamma \frac{1}{z}\,dz-\frac i2\oint_\gamma \frac{1}{z-i/2}\,dz \tag 1 \end{align}$$

Now apply Cauchy's Integral Formula to each of the terms on the right-hand side of $(1)$.

Mark Viola
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