If $f(z) = z^3$ how is it graphed in the sector $0<arg(z)<\frac{\pi}{4}$

Question

If $f(z) = (x+yi)(x^2+ 2xyi - y^2) = x^3 -xy^2 -2xy^2 +i(x^2y-y^3+2x^2y) $ How do I know the way it would be graphed in the sector $0<arg(z)<\frac{\pi}{4}$?

Answer 1

If $z=r(\cos \theta+i\sin \theta),$ then $z^3=r^3(\cos \theta+i\sin \theta)^3=r^3(\cos 3\theta+i\sin 3\theta).$ So the image sector under $z\rightarrow z^3$ is $0<\theta<3\pi/4.$