Using Trig, Represent i + 1 in polar for, and use this to compute (i + 1) squared

Question

Remember use trig to solve this equation, im having a hard time understanding how to even start this so please help.

Answer 1

z=1+i, r=√(1+1)=√2.

z=re^iθ=rcosθ+rsinθ=1+i.

tanθ=1/1=1, θ=π/4, 1+i=√2cos(π/4)+i√2sin(π/4), and z=√2e^iπ/4.

So the polar coordinates are (√2,π/4).

(1+i)²=z²=(√2e^iπ/4)²=2e^iπ/2, which is polar coordinates (2,π/2).

2e^iπ/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.