Question

if f(z)=u(x,y)+iv(x,y) is an analytic function in domain D and f(z) does not equal to 0 for all z in D,show that p(x,y)=log e |f(z)| is harmonic in D.

Answer 1

\[
\frac{\partial ^2f(x,y)}{\partial x^2}+\frac{\partial
^2f(x,y)}{\partial y^2}= \\ \frac{-4 \left(u(x,y)
u^{(0,1)}(x,y)+v(x,y) v^{(0,1)}(x,y)\right)^2-4
\left(u(x,y) u^{(1,0)}(x,y)+v(x,y)
v^{(1,0)}(x,y)\right)^2+\\ 2 \left(u(x,y)^2+v(x,y)^2\right)
\left(u^{(0,1)}(x,y)^2+u(x,y)
u^{(0,2)}(x,y)+v^{(0,1)}(x,y)^2+\\v(x,y)
v^{(0,2)}(x,y)\right)+2 \left(u(x,y)^2+v(x,y)^2\right)
\left(u^{(1,0)}(x,y)^2+\\ u(x,y)
u^{(2,0)}(x,y)+v^{(1,0)}(x,y)^2+v(x,y)
v^{(2,0)}(x,y)\right)}{2
\left(u(x,y)^2+v(x,y)^2\right)^2}\\\text{/.}\,
\left\{u^{(0,1)}(x,y)\to -v^{(1,0)}(x,y),u^{(1,0)}(x,y)\to
v^{(0,1)}(x,y)\right\}

\]

Answer 2

The above is equal to
\[
\frac{u(x,y) \left(u^{(0,2)}(x,y)+u^{(2,0)}(x,y)\right)+v(x,y)
\left(v^{(0,2)}(x,y)+v^{(2,0)}(x,y)\right)}{u(x,y)^2+v(x,y)
^2}=0

\]
Since u and v are harmonic

Answer 3

my f is his p

Answer 4

up to a constant

Answer 5

the brute force way is to take two derivatives of log(a(f(x))), where a is the absolute value function.

for the first derivative df/dx, we get a'(f)f'/a(f)
for the second, we get -(a'(f)f'/a(f))^2+(f'^2 a''(f)+f''a')/a(f)

now to find df/dy, remember df/dy = i df/dx since f is analytic.
Thus the second derivative d2f/dy2 = -(a'(f)*i*f'/a(f))^2+((i*f')^2 a''(f)+i^2*f''a')/a(f) = -d2f/dx2. Thus d2f/dy2+d2f/dx^2=0.

an easier way to solve this is to use the fact that log(|f(x)|)=re(log(f(x))); the real part of any analytic function is harmonic.