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Consider the appropriate contour integral (circle $\oint=e^z$, show that $$\int^{2\pi}_{0}e^{cos\theta}cos(sin\theta +\theta)d\theta = 0$$ A more thorough explanation would be for the better.

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1 Answer 1


$$\oint_{|z|=1} dz\: \exp{z}$$

Apply Cauchy's Theorem to see that this integral is zero. Then parametrize using $z=\exp{(i \theta)}$, $dz = i \exp{(i \theta)}$.

That is,

$$\begin{align}\oint_{|z|=1} dz\: \exp{z} &= i \int_0^{2 \pi} d \theta \: \exp{[\exp{(i \theta)} + i \theta]}\\&=i \int_0^{2 \pi} d \theta \exp{[\cos{\theta} + i (\sin{\theta} + \theta)]} \\ &= i \int_0^{2 \pi} d \theta \: \exp{(\cos{\theta})} [\cos{(\sin{\theta}+\theta)} + i \sin{(\sin{\theta}+\theta)}] \\ \end{align}$$

Since the integral is zero, the real and imaginary parts are each zero.

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could you elaborate a little? – Moses G Feb 5 '13 at 11:30
@Moses G:How is that? – Ron Gordon Feb 5 '13 at 11:39
sorry, didnt see your comment. That's a lot better – Moses G Feb 5 '13 at 12:17

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