math
posted by haritha .
If a+ib = (root of 1+i) / (root of 1i )then prove a^2 +b^2 =1

√(1+i)/√(1i)
= √(1+i)√(1i) / (1i)
= √(1i^2) / (1i)
= √2/(1i)
= √2(1+i) / (1i)(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 
or, using de Moivre's rule,
1+i = √2 cis π/4
so √(1+i) = ∜2 cis π/8
1i = √2 cis π/4
so √(1i) = ∜2 cis π/8
√(1+i)/√(1i) = cis π/4
Now the angle doesn't matter, since cos^2+sin^2 = 1
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