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If $n$ is a positive integer, then $(1+i)^n+(1-i)^n=?$

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  • $(cos\theta+isin\theta)^n=cos\;n\theta+isin\;n\theta$
  • $sin(-\theta)=-sin\theta$
$1+i=\sqrt 2(cos\large\frac{\pi}{4}$$+isin\large\frac{\pi}{4})$
$1-i=\sqrt 2(cos(-\large\frac{\pi}{4})$$+isin(-\large\frac{\pi}{4}))$
$(1+i)^n=(\sqrt 2)^n(cos\large\frac{n\pi}{4}$$+isin\large\frac{n\pi}{4})$
$(1-i)^n=(\sqrt 2)^n(cos(-\large\frac{n\pi}{4})$$+isin(-\large\frac{n\pi}{4}))$
$(1+i)^n+(1-i)^n=(\sqrt 2)^n(2cos\large\frac{n\pi}{4})$
$=2^{\large\frac{n+2}{2}}$$cos\:\large\frac{n\pi}{4}$
answered Jul 31, 2013 by rvidyagovindarajan_1
edited May 26, 2014 by rohanmaheshwari0831_1
 

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