z=1+i, r=√(1+1)=√2.
z=reiθ=rcosθ+rsinθ=1+i.
tanθ=1/1=1, θ=π/4, 1+i=√2cos(π/4)+i√2sin(π/4), and z=√2eiπ/4.
So the polar coordinates are (√2,π/4).
(1+i)2=z2=(√2eiπ/4)2=2eiπ/2, which is polar coordinates (2,π/2).
2eiπ/2=2cos(π/2)+2isin(π/2)=0+2i=2i. So (1+i)2=2i
CHECK
(1+i)2=1+2i-1=2i. This confirms the solution via de Moivre.
When r=2 and θ=π/2, rcosθ=2cos(π/2)=0 (real part is zero) and rsinθ=2sin(π/2)=2.