DeMoivre’s Theorem states, “If z = r(cos u + i sin u), then zn = rn(cos nu + i sin nu). Verify for n=2?

Update:

Ok, the z^2 instead of z2 is really confusing me. Can someone explain why it becomes squared instead of doubled?

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  • 1 decade ago
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    DeMoivre's Theorem allows you to find powers of complex numbers. Given the complex number

    z = r(cos u + i sin u)

    Square both sides,

    z² = r²(cos² u + 2icos u sin u + i²sin² u)

    = r²(cos² u + 2icos u sin u - sin² u)

    = r²[(cos² u - sin² u) + i(2 sin u cos u)]

    = r²(cos 2u + i sin 2u)

    So it's true for n = 2.

  • 1 decade ago

    OK, so plug in n=2.

    z^2 = r^2(cos 2u + i sin 2u)

    That's what you need to verify. Where are you getting stuck?

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