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

> > [!TLDR] > Express $1+i$ and $1-i$ in polar (modulus-amplitude) form, apply De Moivre's theorem to raise each to the $n$-th power, then add. > > Write $1+i$ in polar form: modulus $=\sqrt{1^2+1^2}=\sqrt2$, argument $=\dfrac{\pi}{4}$. > > $1+i = \sqrt2\left(\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4}\right)$ …

If n is a positive integer, then show that (1+i)ⁿ

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Telangana TsbieTelangana Board of Intermediate Education 2019Subjective· 7mImportance★★★★★est

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1\pm i

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

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

Q.If nnn 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)(1+i)n+(1−i)n=22n+2​Cos(4nπ​).

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Q.If nnn 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)(1+i)n+(1−i)n=22n+2​Cos(4nπ​).

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

Unlock the full solution

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

Unlock the full solution