1
$\begingroup$

From my complex analysis text there is an applicative question where we are asked to evaluate the integral; $$ \int_0^{2\pi} e^{i\theta} e^{- i e^{i n \theta}} \: d\theta$$ I figured this could be done in the usual way defining the unit circle as $\Gamma = \{z \in \mathbb{C} \mid |z| =1 \}$, with $z = e^{i \theta}$ and $d \theta = \dfrac{dz}{iz}$. Although I am not sure how to proceed from this setup; $$ - i \oint_\Gamma e^{-i z^n} \: dz $$ Is this function not analytic everywhere in $\Gamma$ and thus by Goursat's theorem $=0$, or have I missed some relevant step?

$\endgroup$
0

1 Answer 1

0
$\begingroup$

Yes $f(z) = e^{-i z^n}$ is analytic on $|z| < 2$ (a simply connected open containing $\Gamma$) so that $\int_{|z| = 1} f(z)dz = 0$.

You can use the Cauchy integral theorem, or simply say that $f(z) = \sum_{k=0}^\infty c_k z^k$ converges uniformly on $|z| < 2$ so that $$\int_{|z| = 1} f(z)dz = \sum_{k=0}^\infty c_k\int_{|z| = 1} z^k dz = 0$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .