Use the Cauchy-Riemann equations to show that the function $f(z) = \exp(\bar{z})$
Here's my attempt:
we have $e^{\bar{z}} = e^{x-iy} = e^{x}e^{-iy} = e^x(\cos(y)-i\sin(y))$
we can then define:
- $u(x,y)= e^x\cos(y)$
- $v(x,y)= -e^x\sin(y)$
Then we can easily see that the Cauchy-Riemann equations aren't satisfied.
What I find weird is that I am asked to show that $f(z)$ is not defined everywhere whereas it's defined almost nowhere.
Is this the right way to go about this problem?