The least positive integer n such that frac{( | vedclass

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

Question

(C) Given expression: $E = \frac{(2i)^{n}}{(1-i)^{n-2}}$First,simplify $(1-i)^2 = 1 + i^2 - 2i = 1 - 1 - 2i = -2i$.Thus,$(1-i)^{n-2} = ((1-i)^2)^{\fra

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Given expression: $E = \frac{(2i)^{n}}{(1-i)^{n-2}}$First,simplify $(1-i)^2 = 1 + i^2 - 2i = 1 - 1 - 2i = -2i$.Thus,$(1-i)^{n-2} = ((1-i)^2)^{\frac{n-2}{2}} = (-2i)^{\frac{n-2}{2}}$.Substituting this into the expression: $E = \frac{(2i)^n}{(-2i)^{\frac{n-2}{2}}} = \frac{(2i)^n}{(-1)^{\frac{n-2}{2}} (2i)^{\frac{n-2}{2}}} = \frac{(2i)^{n - \frac{n-2}{2}}}{(-1)^{\frac{n-2}{2}}} = \frac{(2i)^{\frac{n+2}{2}}}{(-1)^{\frac{n-2}{2}}}$.For $E$ to be a positive integer,the power of $i$ must be a multiple of $4$ (i.e.,$\frac{n+2}{2} = 4k$) and the magnitude must be real.If $n=6$,$\frac{n+2}{2} = 4$. Then $E = \frac{(2i)^4}{(-1)^2} = \frac{16 i^4}{1} = 16(1) = 16$,which is a positive integer.Thus,the least positive integer $n$ is $6$.

The least positive integer $n$ such that $\frac{(2i)^{n}}{(1-i)^{n-2}}$,where $i=\sqrt{-1}$,is a positive integer,is ..... .

Similar Questions

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

If $\frac{3+2i \sin \theta}{1-2i \sin \theta}$ is a real number and $0 < \theta < 2\pi$,then $\theta$ is equal to

If $\left(\frac{1-i}{1+i}\right)^{100}=a+ib$,where $a, b \in \mathbb{R}$ and $i=\sqrt{-1}$,then $(a, b)$ is equal to

$\left( \frac{1}{1 - 2i} + \frac{3}{1 + i} \right) \left( \frac{3 + 4i}{2 - 4i} \right) = $

The imaginary part of $\frac{(1 + i)^2}{2 - i}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module