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Show that, (1+i)n+(1i)n=2n+22cosnπ4
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Solution:

Let, 1+i=r(cosθ+isinθ)=rcosθ+isinθ Equating real \& imaginary parts on both sides,

rcosθ=1&rsinθ=1

Now,

(rcosθ)2+(rsinθ)2=(1)2+(1)2

Now,

(rcosθ)2+(rsinθ)2=(1)2+(1)2

r2cos2θ+r2sin2θ=1+1

r2(cos2θ+sin2θ)=2

r2(1)=2

r2=2r=2

Also, rsinθrcosθ=11

tanθ=1θ=tan1(1)θ=π41+i=2[cosπ4+isinπ4]

Similarly we can prove that, 1i=2[cosπ4isinπ4]

LHS: (1+i)n+(1i)n

=[2(cosπ4+isinπ4]n+[2(cosπ4isinπ4)]n=(2)n[cosπ4+isinπ4]n+(2)n[cosπ4isinπ4]n=(2)n[cosnπ4+isinnπ4+cosnπ4isinnπ4]=2n22cosnπ4=2n2+1cosnπ4=2n+22cosnπ4= RHS 

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