The least positive integer n,for which frac{( | vedclass

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

Question

(C) Given expression: $\frac{(1+i)^{n}}{(1-i)^{n-2}}$$= \frac{(1+i)^{n}}{(1-i)^{n}} 	imes (1-i)^{2}$$= \left(\frac{1+i}{1-i}ight)^{n} 	imes (1 + i

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given expression: $\frac{(1+i)^{n}}{(1-i)^{n-2}}$$= \frac{(1+i)^{n}}{(1-i)^{n}} 	imes (1-i)^{2}$$= \left(\frac{1+i}{1-i}ight)^{n} 	imes (1 + i^{2} - 2i)$$= \left(\frac{(1+i)(1+i)}{(1-i)(1+i)}ight)^{n} 	imes (1 - 1 - 2i)$$= \left(\frac{1 + i^{2} + 2i}{1 - i^{2}}ight)^{n} 	imes (-2i)$$= \left(\frac{2i}{2}ight)^{n} 	imes (-2i)$$= i^{n} 	imes (-2i) = -2i^{n+1}$For the expression to be a positive real number,$-2i^{n+1}$ must be positive.If $n=1$,$-2i^{1+1} = -2i^{2} = -2(-1) = 2$,which is positive.Thus,the least positive integer $n$ is $1$.

The least positive integer $n$,for which $\frac{(1+i)^{n}}{(1-i)^{n-2}}$ is positive,is

Similar Questions

${\left( \frac{1 + i}{1 - i} \right)^2} + {\left( \frac{1 - i}{1 + i} \right)^2}$ is equal to

$\frac{1 - 2i}{2 + i} + \frac{4 - i}{3 + 2i} = $

Solve $x^{2}+2=0$.

If $z = 1 + i,$ then the multiplicative inverse of $z^2$ is (where $i = \sqrt{-1}$)

The least positive integer $n$ which will reduce $\left( \frac{i - 1}{i + 1} \right)^n$ to a real number is

Vedclass Test Series

Exam Paper Generator

Online Exam Module