Cauchy-Riemann Equations

Question

I have a question. If I have a complex function and I know that for $r=1$, $f(z=e^{iθ})=\cos(θ)\sin(θ)$ And I know that $f(z)$ is entire. Now, I need to calculate $f'(z)$ for every $z$ in $r=1$, meaning $z=e^{iθ}$. I tried to define $u(r,θ)= \cos(θ)\sin(θ)$, $v(r,θ)= 0$ And use Cauchy-Riemann Equations in polar coordinates..and then I get that the derivative is zero, but I am not sure cause I can calculate the derivative by definition (derivative by $θ$) and then I got something which is a function of $θ$ so I am a bit confused..why is that? What is the right way to calculate it?

Answer 1

$f(z)=\frac 1 {4i} (z^{2}-\frac 1 {z^{2}})$ whenever $|z|=1$. By Identity Theorem we must have $f(z)=\frac 1 {4i} (z^{2}-\frac 1 {z^{2}})$ for all $z \neq 0$, so $f$ cannot be continuous at $0$. Hence, there is no such entire function.

If you consider $f$ as an analytic function on $\mathbb C \setminus \{0\}$ then we can compute $f'(z)$ easily: $f'(z)=\frac 1 {4i} (2z+\frac 2 {z^{3}})$. This is valid for all $z \neq 0$, in particular for $|z|=1$.