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Q.If nn is a positive integer, then show that (1+i)n+(1−i)n=2n+22Cos(nπ4)(1+i)^n + (1-i)^n = 2^{\frac{n+2}{2}} Cos\left(\dfrac{n\pi}{4}\right).

Telangana TsbieTelangana Board of Intermediate Education 2019Subjective· 7mImportance★★★★★
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Express 1+i1+i and 1−i1-i in polar (modulus-amplitude) form, apply De Moivre's theorem to raise each to the nn-th power, then add.

Write 1+i1+i in polar form: modulus =12+12=2=\sqrt{1^2+1^2}=\sqrt2, argument =π4=\dfrac{\pi}{4}.

1+i=2(cos⁡π4+isin⁡π4)1+i = \sqrt2\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)

Similarly, 1−i=2(cos⁡π4−isin⁡π4)=2(cos⁡(−π4)+isin⁡(−π4))1-i = \sqrt2\left(\cos\dfrac{\pi}{4}-i\sin\dfrac{\pi}{4}\right) = \sqrt2\left(\cos\left(-\dfrac{\pi}{4}\right)+i\sin\left(-\dfrac{\pi}{4}\right)\right).

By De Moivre's theorem, (cos⁡ϕ+isin⁡ϕ)n=cos⁡nϕ+isin⁡nϕ(\cos\phi+i\sin\phi)^n = \cos n\phi + i\sin n\phi:

(1+i)n=(2)n(cos⁡nπ4+isin⁡nπ4)(1+i)^n = (\sqrt2)^n\left(\cos\dfrac{n\pi}{4}+i\sin\dfrac{n\pi}{4}\right)

(1−i)n=(2)n(cos⁡nπ4−isin⁡nπ4)(1-i)^n = (\sqrt2)^n\left(\cos\dfrac{n\pi}{4}-i\sin\dfrac{n\pi}{4}\right)

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