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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

$du/dx=-\sin(x)\cosh(y)$ $dv/dy=-\sin(x)\cosh(y)$

$du/dy=\cos(x)\sinh(y)$ $dv/dx=-\cos(x)\sinh(y)$

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    @ChrisEagle sorry forgot to tag you in my comment2012-11-26

3 Answers 3

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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.

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    @TheLastCipher If $z \mapsto f(z)$ is an analytic function, then $z \mapsto c f(z)$ is analytic for any constant $c \in \mathbb{C}$. Since $z \mapsto e^{iz}+e^{-iz}$ is analytic (as sum of two analytic functions) we get that $z \mapsto 1/(2i) (e^{iz}+e^{-iz})$ is analytic.2018-07-30
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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.