1
326views
Show that, (1+i)n+(1−i)n=2n+22cosnπ4.
1 Answer
written 3.0 years ago by | • modified 3.0 years ago |
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=2⇒r=√2
Also, rsinθrcosθ=11
⇒tanθ=1⇒θ=tan−1(1)⇒θ=π4∴1+i=√2[cosπ4+isinπ4]
Similarly we can prove that, 1−i=√2[cosπ4−isinπ4]
LHS: (1+i)n+(1−i)n
=[√2(cosπ4+isinπ4]n+[√2(cosπ4−isinπ4)]n=(√2)n[cosπ4+isinπ4]n+(√2)n[cosπ4−isinπ4]n=(√2)n[cosnπ4+isinnπ4+cosnπ4−isinnπ4]=2n22cosnπ4=2n2+1cosnπ4=2n+22cosnπ4= RHS