DeMoivre's Theorem: Standard FormDate: 3/19/96 at 18:43:13 From: Robert Holscher Subject: Math Dr. Math, I have no idea how to begin to solve this problem. Use DeMoivre's theorem to write (1-i)^10 in standard form. Please help! Rob Holscher Date: 3/22/96 at 19:23:50 From: Doctor Sebastien Subject: Re: Math Hi, De Moivre's Theorem: (cos x + i sin x)^n = cos nx +i sin nx, where n is any real number. The key to your problem is to find x such that cos x = 1 and sin x = -1. When x =- Pi/4, cos x=1/SQRT(2) and sin x = -1/SQRT(2), where SQRT means square root. Therefore, (1-i)^10 = {SQRT(2)*(cos (-Pi/4) + i sin (-Pi/4))}^10 = (SQRT(2)^10)*(cos (-5Pi/2) + i sin (-5Pi/2)), by DeMoivre's theorem =32 * -i = -32i Doctor Sebastien, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/