# Find function value outside definition domain

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?

• 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. – elescararriba Jun 15 at 7:38
• But how could I extend the function to the whole complex plane plane? – merlo94 Jun 15 at 8:29
• Identical question shared here: math.stackexchange.com/a/3263246/682282 – merlo94 Jun 15 at 13:33
• 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. – Andreas Blass Jun 15 at 15:07