Is it possible to prove that the gradients of the real and imaginary parts of a complex analytic functions have the same length? Suppose I have a complex analytic function 
$$f(x,y)=(x+iy)^n$$ 
where both $x$ and $y$ are real and $n$ is an integer. Is it possible to prove that the gradient of the real part of $f$ and the gradient of the imaginary part of $f$ are the same length for any $n$? 
I've tried several different values of n and in each example the gradients for the real and imaginary parts had the same length.
So far I've attempted to use the binomial theorem to expand the function $f$ out, in the hope that I might spot some pattern but this hasn't really helped. Can anyone offer any suggestions or proof?
Any help greatly appreciated.
Stephen
 A: If the function is "complex differentiable", or "analytic", then your statement is true. Writing 
$$
f(x + iy) = u(x, y) + i v(x, y)
$$
you're asking whether 
$$
\left( \frac{\partial u} {\partial x}  \right)^2 + \left( \frac{\partial u} {\partial y}  \right)^2  = 
\left( \frac{\partial v} {\partial x}  \right)^2 + \left( \frac{\partial v} {\partial y}  \right)^2.
$$
This equality is guaranteed by the so-called "Cauchy Riemann equations", which say that 
$$
\frac{\partial u} {\partial x} = \frac{\partial v} {\partial y} \\
\frac{\partial u} {\partial y} = -\frac{\partial v} {\partial x},
$$
for any complex-differentiable function. 
For a non-differentiable function, your claim is false. For instance, if you write 
$$
f(x + iy) = x + 0i
$$
then the length of the gradient of the real part is $1$, but the length of the gradient of the complex part is 0. 
A: If $f$ is analytic and $f = u+iv$ you have the Cauchy-Riemann equations $u_x = v_y$ and $u_y = -v_x$. Since $$|\nabla u|^2 = u_x^2 + u_y^2,\quad |\nabla v|^2 = v_x^2 + v_y^2$$ it's not hard to see $|\nabla u| = |\nabla v|$.
