0
$\begingroup$

An old exam style question I have encountered asks to evaluate the integral of the following;

$\int_{C}\frac{1}{1+\sqrt{z+2}}$ where $C$ is the positively oriented unit circle.

So firstly I proceeded to parametrise the integral, using the fact that the unit circle $C=e^{it}$ and therefore $C'=ie^{it}$. However thought Cauchys Theorem could provide an easier solution, as the function is analytic in the whole complex plane and our path is contained inside the complex plane, therefore by Cauchy's would equal zero. Am i correct, or I have I swerved completely off track with my train of thought ?

$\endgroup$

1 Answer 1

1
$\begingroup$

The function is not analytic on the whole plane. If $\sqrt{\ }$ is taken to be the principal branch, $\sqrt{2+z}$ is analytic on $D=\mathbb{C}\setminus(-\infty,-2]$. Then $1+\sqrt{2+z}\ne0$ for all $z\in D$ and the integrand is analytic in $D$. Since $C\subset D$, it follows that the integral is $0$.

$\endgroup$
1
  • $\begingroup$ Of course! Thank you. $\endgroup$
    – user112365
    Apr 22, 2015 at 13:50

You must log in to answer this question.

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