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.


Leave a Comment »

No comments yet.

RSS feed for comments on this post. TrackBack URI

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

Blog at

%d bloggers like this: