If a+ib = (root of 1+i) / (root of 1-i )then prove a^2 +b^2 =1

Answer 1

√(1+i)/√(1-i)
= √(1+i)√(1-i) / (1-i)
= √(1-i^2) / (1-i)
= √2/(1-i)
= √2(1+i) / (1-i)(1+i)
= √2(1+i)/2
= 1/√2 + 1/√2 i

a = 1/√2
b = 1/√2
a^2+b^2 = 1/2 + 1/2 = 1

Answer 2

or, using de Moivre's rule,

1+i = √2 cis π/4
so √(1+i) = ∜2 cis π/8

1-i = √2 cis -π/4
so √(1-i) = ∜2 cis -π/8

√(1+i)/√(1-i) = cis π/4

Now the angle doesn't matter, since cos^2+sin^2 = 1