Let f(z) be continuous on D, a domain.
Prove that f(z) is analytic <=> (f(z))^2 is analytic.
(=>) is easy.
(<=) is problematic.
f(z)=u+iv
so f(z)^2=(u^2-v^2)+i(2uv)
The Cauchy Riemann Equations imply
u*u_x - v*v_x = v*u_y + u*v_y
u*u_y - v*v_y =-v*u_x + u*v_x
where _x denotes the partial with respect to x.
Now there are problems. Solving for u or v, you wind up dividing by something you wish to prove is 0.
now let:
A = u_x - v_y
B = u_y + v_x
we get:
uB = -vA
uA = vB
multiply both sides by u for the first equation, v for the second.
Get Au^2=-Av^2
Which is impossible unless A is zero (what we want to prove) OR u, v or both = 0. I tried to solve these cases, but didn't get anywhere