How to deduce the analyticity of a function, $f(z) = \sin (2z)$

Question

Deduce the analyticity of the function $ f(z) = \sin (2z) $.

Answer 1

If $x,y\in\mathbb R$, then$$\sin\bigl(2(x+yi)\bigr)=\overbrace{\sin(2x)\cosh(2y)}^{=u(x,y)}+\overbrace{\cos(2x)\sinh(2y)}^{=v(x,y)}i.$$It is easy to check that both partial derivatives of both functions $u$ and $v$ are continuous everywhere.

Answer 2

If Cauchy Riemann equations are satisfied at every point of an open set then the function is analytic in that set. [But if Cauchy Riemann equations are satisfied at a particular point you cannot say that the function is analytic at that point].

Answer 3

One way is using $\sin2z=\dfrac{1}{2i}\left(e^{2iz}-e^{-2iz}\right)$ where $e^z$ is entire.

Answer 4

For all $z\in\mathbb{C}$ $$\sin(2z)=(2z)-\frac{(2z)^3}{3!}+\frac{(2z)^5}{5!}+...=\sum_{n=0}^\infty\frac{(-1)^n2^{2n+1}}{(2n+1)!}z^{2n+1}$$ and because radius of convergence of this series is infinite $f$ is analytic on the whole complex plane.

Or if you already know that $\sin z$ is analytic, then because $z\mapsto2z$ is also analytic $f$ is also analytic in $\mathbb{C}$ as a composition of analytic functions.