If n is a positive integer,then (1 + i)^n + ( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We express $1+i$ and $1-i$ in polar form: $1+i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$ $1-i = \sqrt{2} \left( \cos \f

Vedclass · Accepted Answer

$(\sqrt{2})^{n+2} \cos \left( \frac{n\pi}{4} \right)$

Vedclass · Answer

(C) We express $1+i$ and $1-i$ in polar form: $1+i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right)$ $1-i = \sqrt{2} \left( \cos \frac{\pi}{4} - i \sin \frac{\pi}{4} \right)$ Using De Moivre's theorem: $(1+i)^n + (1-i)^n = (\sqrt{2})^n \left( \cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4} \right) + (\sqrt{2})^n \left( \cos \frac{n\pi}{4} - i \sin \frac{n\pi}{4} \right)$ $= 2 \cdot (\sqrt{2})^n \cos \frac{n\pi}{4}$ $= 2^1 \cdot 2^{n/2} \cos \frac{n\pi}{4}$ $= 2^{(n/2) + 1} \cos \frac{n\pi}{4}$ $= (\sqrt{2})^{n+2} \cos \frac{n\pi}{4}$

If $n$ is a positive integer,then $(1 + i)^n + (1 - i)^n$ is equal to

Similar Questions

Let $z = \frac{1 - i \sqrt{3}}{2}$,where $i = \sqrt{-1}$. Then the value of $21 + \left(z + \frac{1}{z}\right)^{3} + \left(z^{2} + \frac{1}{z^{2}}\right)^{3} + \left(z^{3} + \frac{1}{z^{3}}\right)^{3} + \dots + \left(z^{21} + \frac{1}{z^{21}}\right)^{3}$ is .... .

If $x+\frac{1}{x}=2 \cos \theta$, then for any integer $n$, $x^n+\frac{1}{x^n}=$

If $2+2 \sqrt{3} i=k(\cos \theta+i \sin \theta)$ where $k > 0$,then find the value of $\frac{1}{\sqrt{3}}[\cos 6 \theta+i \sin 6 \theta]$.

If $\omega$ is an imaginary cube root of unity,then the value of $(1-\omega+\omega^{2}) \cdot(1-\omega^{2}+\omega^{4}) \cdot(1-\omega^{4}+\omega^{8}) \cdot \ldots$ ($2n$ factors) is

If $\omega \neq 1$ is a cube root of unity,then one root among the $7^{\text{th}}$ roots of $(1+\omega)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module