# College Math Teaching

## May 3, 2012

### Composing a non-constant analytic function with a non-analytic one, part II

Filed under: advanced mathematics, analysis, calculus, complex variables, matrix algebra — collegemathteaching @ 6:40 pm

I realize that what I did in the previous post was, well, lame.
The setting: let $g$ be continuous but non-analytic in some disk $D$ in the complex plane, and let $f$ be analytic in $g(D)$ which, for the purposes of this informal note, we will take to contain an open disk. If $g(D)$ doesn’t contain an open set or if the partials of $g$ fail to exist, the question of $f(g)$ being analytic is easy and uninteresting.

Let $f(r + is ) = u(r,s) + iv(r,s)$ and $g(x+iy) = r(x,y) + is(x,y)$ where $u, v, r, s$ are real valued functions of two variables which have continuous partial derivatives. Assume that $u_r = v_s$ and $u_s = -v_r$ (the standard Cauchy-Riemann equations) in the domain of interest and that either $r_x \neq s_y$ or $r_y \neq -s_x$ in our domain of interest.

Now if the composition $f(g)$ is analytic, then the Cauchy-Riemann equations must hold; that is:
$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$

Now use the chain rule and do some calculation:
From the first of these equations:
$u_r r_x + u_s s_x = v_r r_y + v_s s_y$
$u_r r_y + u_s s_y = -v_r r_x - v_s s_x$
By using the C-R equations for $u, v$ we can substitute:
$u_r r_x + u_s s_x = -u_s r_y + u_r s_y$
$u_r r_y + u_s s_y = u_s r_x - u_r s_x$
This leads to the following system of equations:
$u_r(r_x -s_y) + u_s(s_x + r_y) = 0$
$u_r(r_y + s_x) + u_s(s_y - r_x) = 0$
This leads to the matrix equation:
$\left( \begin{array}{cc}(r_x -s_y) & (s_x + r_y) \\(s_x + r_y) & (s_y - r_x) \end{array} \right)\ \left(\begin{array}{c}u_r \\u_s \end{array}\right)\ = \left(\begin{array}{c} 0 \\ 0 \end{array}\right)\$

The coefficient matrix has determinant $-((r_x - s_y)^2 + (s_x + r_y)^2)$ which is zero when BOTH $(r_x - s_y)$ and $(s_x + r_y)$ are zero, which means that the Cauchy-Riemann equations for $g$ hold. Since that is not the case, the system of equations has only the trivial solution which means $u_r = u_s = 0$ which implies (by C-R for $f$ ) that $v_r = v_s = 0$ which implies that $f$ is constant.

This result includes the “baby result” in the previous post.