# College Math Teaching

## May 2, 2012

### Composition of an analystic function with a non-analytic one

Filed under: advanced mathematics, analysis, complex variables, derivatives, Power Series, series — collegemathteaching @ 7:39 pm

On a take home exam, I gave a function of the type: $f(z) = sin(k|z|)$ and asked the students to explain why such a function was continuous everywhere but not analytic anywhere.

This really isn’t hard but that got me to thinking: if $f$ is analytic at $z_0$ and NON CONSTANT, is $f(|z|)$ ever analytic? Before you laugh, remember that in calculus class, $ln|x|$ is differentiable wherever $x \neq 0$.

Ok, go ahead and laugh; after playing around with the Cauchy-Riemann equations at bit, I found that there was a much easier way, if $f$ is analytic on some open neighborhood of a real number.

Since $f$ is analytic at $z_0$, $z_0$ real, write $f = \sum ^ {\infty}_{k =0} a_k (z-z_0)^k$ and then compose $f$ with $|z|$ and substitute into the series. Now if this composition is analytic, pull out the Cauchy-Riemann equations for the composed function $f(x+iy) = u(x,y) + iv(x,y)$ and it is now very easy to see that $v_x = v_y =0$ on some open disk which then implies by the Cauchy-Riemann equations that $u_x = u_y = 0$ as well which means that the function is constant.

So, what if $z_0$ is NOT on the real axis?

Again, we write $f(x + iy) = u(x,y) + iv(x,y)$ and we use $U_{X}, U_{Y}$ to denote the partials of these functions with respect to the first and second variables respectively. Now $f(|z|) = f(\sqrt{x^2 + y^2} + 0i) = u(\sqrt{x^2 + y^2},0) + iv(\sqrt{x^2 + y^2},0)$. Now turn to the Cauchy-Riemann equations and calculate:
$\frac{\partial}{\partial x} u = u_{X}\frac{x}{\sqrt{x^2+y^2}}, \frac{\partial}{\partial y} u = u_{X}\frac{y}{\sqrt{x^2+y^2}}$
$\frac{\partial}{\partial x} v = v_{X}\frac{x}{\sqrt{x^2+y^2}}, \frac{\partial}{\partial y} v = v_{X}\frac{y}{\sqrt{x^2+y^2}}$
Insert into the Cauchy-Riemann equations:
$\frac{\partial}{\partial x} u = u_{X}\frac{x}{\sqrt{x^2+y^2}}= \frac{\partial}{\partial y} v = v_{X}\frac{y}{\sqrt{x^2+y^2}}$
$-\frac{\partial}{\partial x} v = -v_{X}\frac{x}{\sqrt{x^2+y^2}}= \frac{\partial}{\partial y} u = u_{X}\frac{y}{\sqrt{x^2+y^2}}$

From this and from the assumption that $y \neq 0$ we obtain after a little bit of algebra:
$u_{X}\frac{x}{y}= v_{X}, u_{X} = -v_{X}\frac{x}{y}$
This leads to $u_{X}\frac{x^2}{y^2} = v_{X}\frac{x}{y}=-v_{X}$ which implies either that $u_{X}$ is zero which leads to the rest of the partials being zero (by C-R), or this means that $\frac{x^2}{y^2} = -1$ which is absurd.

So $f$ must have been constant.