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Q.If nn is an integer, then show that (1+i)2n+(1−i)2n=2n+1cos⁡(nπ2)(1 + i)^{2n} + (1 - i)^{2n} = 2^{n+1} \cos\left(\frac{n\pi}{2}\right).

Telangana TsbieTelangana Board of Intermediate Education 2025Subjective· 7mImportance★★★★★
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1±i=2 Cis(±π/4)1\pm i=\sqrt2\,\text{Cis}(\pm\pi/4); De Moivre gives 2nCis(±nπ/2)2^n\text{Cis}(\pm n\pi/2), whose sum is 2n+1cos⁡(nπ/2)2^{n+1}\cos(n\pi/2).

In polar form,

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

By De Moivre's theorem,

(1+i)2n=(2)2n(cos⁡2nπ4+isin⁡2nπ4)=2n(cos⁡nπ2+isin⁡nπ2),(1+i)^{2n}=(\sqrt2)^{2n}\big(\cos\tfrac{2n\pi}{4}+i\sin\tfrac{2n\pi}{4}\big)=2^{n}\big(\cos\tfrac{n\pi}{2}+i\sin\tfrac{n\pi}{2}\big),

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