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?
Best Answer
First, we want to find the partial derivative of $u(x, y)$ and $v(x, y)$ with respect to both $x$ and $y$.
$$u_x = e^x \cos(y)$$ $$u_y = -e^x \sin(y)$$ $$v_x = -e^x \sin(y)$$ $$v_y = -e^x \cos(y)$$
The Cauchy-Riemann equations are satisfied when:
$$v_x = v_y$$ $$v_y = -v_x$$
As you pointed out, these equations are clearly not satisfied. A function $f(z)$ can only be analytic at a point if the Cauchy-Riemann equations are satisfied at that point. If they are not, then the function is not analytic there. Thus, because there are points where $f(z)$ is not analytic, it is not analytic everywhere. Hope that answers your question.