# Value of complex integral

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?

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

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