Verify that 1+i = √2 cos 45° + i sin 45° , and compute (1+i)^100?

Answer 1

1 + i = √2 (cos 45° + i sin 45°)

Using De Moivre’s theorem:

If z =  r(cos θ + i sin θ) then zⁿ = [r(cos θ + i sin θ)]ⁿ =[rⁿ(cos nθ + i sin nθ)]

(1 + i )¹⁰ = [√2 (cos 45° + i sin 45°)]¹⁰

              = (√2 )¹⁰(cos 10(45°) + i sin10(45°))

              =(√2)¹⁰(cos 450° + i sin 450°)

              =(2)⁵(cos (360°+90°) + i sin(360°+90°))

               =(2)⁸(cos 90° + i sin 90°)

               =256(0 + i (1))

               =256 i