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