0
$\begingroup$

Given the analytic function $f(z) = u(x,y) + iv(x,y)$, given that $$ u(x,0) = \sin^2x \ , \ v(x,0) = 0 $$ Find $f(z_0)$ for $z_0 = 5 + i8$.

From Cauchy integral representation I could say that $$ f(z_0) = \frac{1}{2 \pi i} \int_{\gamma} \frac{f(z)}{z-z_0}dz $$ with $\gamma$ = $\gamma_1$ + $\gamma_2$ and $$ \gamma_1 = \mathbb{R} \ , \ \gamma_2 = Re^{i\theta} \ \text{for} \ -\pi \leq \theta \leq + \pi $$
Thus I am able to evaluate $g(z) = \dfrac{f(z)}{z-z_0}$ over $\gamma_1$, but not over $\gamma_2$. How could I proceed?

$\endgroup$
  • $\begingroup$ I would say that you can prove that $f(z)=sin^2(z)$ and I'd say that the idea to prove that is to write $f$ as a power series and check that the coefficients must coincide. Then you can solve it much easier. $\endgroup$ – elescararriba Jun 15 at 7:38
  • $\begingroup$ But how could I extend the function to the whole complex plane plane? $\endgroup$ – merlo94 Jun 15 at 8:29
  • $\begingroup$ Identical question shared here: math.stackexchange.com/a/3263246/682282 $\endgroup$ – merlo94 Jun 15 at 13:33
  • $\begingroup$ You don't need to "extend the function"; you're given that $f$ is a function of a complex variable. What needs to be extended is the available information about the function, namely that it agrees with $\sin^2$ on the real axis. For that, use the principle that, if two analytic functions on a domain agree on a set that has a limit point in that domain, then they agree throughout that domain. $\endgroup$ – Andreas Blass Jun 15 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.