# Use the Cauchy-Riemann equations to show that the function $f(z) = \exp(\bar{z})$ is not analytic everywhere.

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.

• $f(z)$ is defined everywhere! – Nosrati Oct 2 '17 at 18:40

First, we want to find the partial derivative of $$u(x, y)$$ and $$v(x, y)$$ with respect to both $$x$$ and $$y$$.
$$u_x = e^x \cos(y)$$ $$u_y = -e^x \sin(y)$$ $$v_x = -e^x \sin(y)$$ $$v_y = -e^x \cos(y)$$
$$v_x = v_y$$ $$v_y = -v_x$$
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.
• Well, if you instead define z conjugate = x - iy, and then take the partial derivatives with respect to that, you get $u_x = 1$ and $v_y = -1$, which is true nowhere. – user484876 Oct 2 '17 at 18:53