[Math] Show that the function, $f(z)=e^{y}e^{ix}$ is nowhere analytic.

complex-analysis

$f(z)=e^{y}e^{ix}$

$=e^{y}(\cos x+i\sin x)$

$u(x,y)=e^{y} \cos x$

$v(x,y)=e^{y}\sin x$

$\frac{\partial u}{\partial x}=-e^{y} \sin x$

$\frac{\partial v}{\partial x}=e^{y} \cos x$

$\frac{\partial u}{\partial y}=e^{y}\cos x$

$\frac{\partial v}{\partial y}=e^{y}\sin x$

We see that the Cauchy-Riemann equations are not satisfied anywhere in the z-plane except at $z=n\pi$ where n is an integer.

We need to check if $f^{'}(z)$ exists at $z=n\pi$

$f^{'}(z)=Lt _{\delta z -> 0}\frac{f(z+\delta z)-f(z)}{\delta z}$

$\delta z = \delta x + i \delta y$

Best Answer

Hint: $z=x+iy$ and

$$f(z)=e^{i\bar{z}}.$$

The C.R. equations are equivalent to the single equation...