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Detemine the value of the integral $$ \int\limits_\gamma e^{4z+1}dz $$

where $\gamma$ is any circle of radius $\frac{1+\sqrt{5}}{2}$ in $\mathbb{C}$, oriented counter-clockwise. (5 marks)

I can't see any singularities here, so does that mean I can use Cauchy's theorem on this?

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2  
Function under integral belongs to $\mathcal{O}(\mathbb{C})$, so... –  Norbert Apr 7 '12 at 9:00
3  
The "any circle" bit should be a hint that you indeed use $\oint_\gamma fdz=0$. –  anon Apr 7 '12 at 9:04

1 Answer 1

up vote 3 down vote accepted

You know that this integral should be zero by Cauchy's integral theorem but I suppose they want you to compute it. So here it goes:

Set $r := \frac{1 + \sqrt{5}}{2}$ and $\gamma (t) := r e^{i t}$. Then $$ \oint f(z) dz = \int_0^{2 \pi} f(\gamma) \dot{\gamma}(t) dt = \int_0^{2 \pi} e^{4r (\cos t + i \sin t) + 1} r(i \cos t - \sin t) dt $$

Then $$ \frac{d}{dt}\left ( e^{4r (\cos t + i \sin t) + 1} \right ) = e^{4r (\cos t + i \sin t) + 1} 4r(i \cos t - \sin t)$$

So $$ \oint f(z) dz = \left [ \frac{1}{4} e^{4r (\cos t + i \sin t) + 1} \right ]_0^{2 \pi} = \left [ \frac{1}{4} e^{4r e^{it} + 1} \right ]_0^{2 \pi} = 0 $$

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