# Complex Analysis Q

Just wondering if this is right, or the right approach as I've not done it before!

Evaluate: $$\int_{|z|=1} \frac {\sin(z)}z dz$$

I expanded $\sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} + \cdots$

So $\frac{\sin(z)}{z} = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} + \cdots$ Which left an integral of: $$\int_{|z|=1}\sum_{n=0}^{\infty} \frac{(-1)^nz^{2n}}{(n+1)!} dz$$

From the unit circle, $z=e^{i\theta} \Rightarrow dz = ie^{i\theta} d\theta$ which gives:

$$\sum_{n=0}^{\infty}\frac {(-1)^ni}{(n+1)!}\int_{0}^{2\pi} e^{3ni\theta} d\theta$$ $$= \sum_{n=0}^{\infty} \frac {(-1)^n(e^{6\pi ni}-1)}{3n(n+1)!}$$ $$= \sum_{n=1}^{\infty} \frac {-2(-1)^n}{3n(n+1)!}$$

Bit unsure if that's right, or the best approach if it is. Thanks!

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But note that $e^{3ni2\pi} = e^ {6\pi n i} = 1$, not $-1$, unless $n=0$, and that $e^{3ni(0)} = 1$ always. So the answer is much simpler than what you written in the end.
@Brit: You should take a look at the $n=0$ term a little bit closer. – mixedmath Mar 18 '13 at 10:09
@Brit Integrating $e^{0}$ from $0$ to $2\pi$ does not give zero, and $(-1)^0i/(0+1)!$ is also not zero. So you oversimplified your integral a little bit. – mixedmath Mar 18 '13 at 10:11
Oh, I see - staring me in the face! So simply $2\pi i$ – Mike Miller Mar 18 '13 at 10:35