Hi guys: just needed some verification on my working out for the following problem:

Question:

Using Cauchy Riemann prove that the function h(z)=sin(Imz) is not differentiable at any point of the strip $\displaystyle \{ z:\frac{{ - \pi }}

{2} < {Im} z < \frac{\pi }

{2}\}$

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My Solution: Please check to see whether it is correct or not.

let $\displaystyle z=x+iy$

$\displaystyle h(z)=\sin (Im \,\, z) = \sin y$

$\displaystyle u(x,y)= \sin y$ and $\displaystyle v(x,y)=0$

$\displaystyle \frac{\delta u}{\delta x}=0$ ... $\displaystyle \frac{\delta v}{\delta x}=0$

$\displaystyle \frac{\delta u}{\delta y}=cos y$, ... $\displaystyle \frac{\delta v}{\delta y}=0$

for CR conditions to be fulfilled, $\displaystyle u_x = v_y$ and $\displaystyle v_x = -u_y$

0 = 0 but

$\displaystyle 0 \neq -cos y$

both the CRE conditions aren't fulfilled so the function is not differentiable at any point of the strip

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is this correct?

thanks