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QUESTION
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\item[(a)]
Show that $f(z)=|z|^2$ is not analytic anywhere.

\item[(b)]
Show that $f(z)=e^z$ is analytic everywhere.

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ANSWER
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\item[(a)]
$z=x+iy,\ |z|^2=x^2+y^2,\ u=x^2,\ v=y^2\\ u_x=2x \neq v_y=2y$
Hence not analytic.

\item[(b)]
$e^z=e^{x+iy}=e^x(\cos y+i \sin y),\ u=e^x\cos y,\ v=e^x \sin y\\
u_x=e^x \cos y =v_y,\ u_y=-e^x\sin y =-v_x$

The partial derivatives are continuous and hence the function is 
analytic.

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