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.

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1 Comment »

  1. […] realize that what I did in the previous post was, well, lame. The setting: let be continuous but non-analytic in some disk in the complex […]

    Pingback by Composing a non-constant analytic function with a non-analytic one, part II « College Math Teaching — May 3, 2012 @ 6:40 pm


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