Show that $\sin(z)$ is analytic I have show that the function $\sin(z)$ satisfies the Cauchy-Riemann equations but don't know where to go from here.
If it saves some working for you they are
$$du/dx=-\sin(x)\cosh(y)$$          $$dv/dy=-\sin(x)\cosh(y)$$
$$du/dy=\cos(x)\sinh(y)$$            $$dv/dx=-\cos(x)\sinh(y)$$
 A: To show $\sin z$ is analytic.
\begin{align}
\sin z&= \sin (x+iy)\\
&= \sin x \cos iy + \cos x\sin iy\\ 
&=\sin x\cosh y + i\cos x \sinh y\\
u&= \sin x \cosh y\\ v&= \cos x \sinh y \\
\frac {∂u}{\partial x} &= \cos x \cosh y\\
\frac {∂u}{∂y}&= \sin x\sinh y\\
\frac {∂v}{∂x}&= -\sin  x \sinh y \\
\frac {∂v}{∂y}&= \cos x \cosh y\\
\frac d{dx} \cosh x &= \sinh x\\
\frac {∂u}{\partial x} &= \frac {∂v} {∂y}\\
 \frac {∂v}{∂x} &= -\frac {∂u}{∂y}\\
\end{align}
Hence the cauchy-riemann equations are satisfied.
Thus $\sin z$ is analytic.
A: By definition we have
$$\sin(z) = \frac{1}{2\imath} \cdot \left(e^{\imath \, z}-e^{-\imath \, z}\right) $$
Since the sum of two analytic functions is analytic, it suffices to show that $z \mapsto e^{\imath \, z}$ and $z \mapsto e^{-\imath \, z}$ are analytic. Let $z:=x+ \imath \, y$ ($x,y \in \mathbb{R}$), then
$$e^{\imath \, z} = e^{\imath \, x} \cdot e^{-y} = \underbrace{e^{-y} \cdot \cos(x)}_{=:u(x,y)}+\imath \, \underbrace{e^{-y} \cdot \sin x}_{=:v(x,y)}$$
From 
$$\partial_x u(x,y) = - e^{-y} \cdot \cos(x) = \partial_y v(x,y) \\
\partial_y u(x,y) = - e^{-y} \cdot \cos(x) = - \partial_x v(x,y)$$
we see that the Cauchy-Riemann equations are satisfied. Since the partial derivatives exist (and are continuous) we conclude that $z \mapsto e^{\imath \, z}$ is analytic. A similar argumentation shows that $z \mapsto e^{-\imath \, z}$ is analytic.
A: $\sin$ and $\cos$ can be defined by the following differential equations


*

*$\cos(0) = 1$

*$\sin(0) = 0$

*$\cos' = -\sin$

*$\sin' = \cos$
It can be shown that there's exactly one way to define $\sin$ and $\cos$ such that they satisfy those differential equations.
Let $f(x) = \sum_{i = 0}^\infty \frac{(-1)^ix^{2i}}{(2i)!}$ and $g(x) = \sum_{i = 0}^\infty \frac{(-1)^ix^{2i + 1}}{(2i + 1)!}$. Both of those series converge on all of $\mathbb{R}$ and it's not that hard to show that every power series that converges on all of $\mathbb{R}$ is analytic on all of $\mathbb{R}$. So $f$ and $g$ are analytic on all of $\mathbb{R}$. It can also be shown that


*

*$f(0) = 1$

*$g(0) = 0$

*$f' = -g$

*$g' = f$
Therefore $f = \cos$ and $g = \sin$ and so $\sin$ and $\cos$ are analytic on all of $\mathbb{R}$.
