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

  • math -

    √(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 -

    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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