Posted by **haritha** on Friday, June 6, 2014 at 8:44pm.

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

- math -
**Steve**, Saturday, June 7, 2014 at 6:15am
√(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

- math -
**Steve**, Saturday, June 7, 2014 at 11:59am
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

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