The value of {i^{22}-(frac{1}{i})^{35}}^2 is | vedclass

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

Question

(A) We know that $i^2 = -1$,$i^4 = 1$. First,calculate $i^{22}$: $i^{22} = (i^4)^5 	imes i^2 = 1^5 	imes (-1) = -1$. Next,calculate $(\frac{1}{i})^{

Vedclass · Accepted Answer

$2i$

Vedclass · Answer

(A) We know that $i^2 = -1$,$i^4 = 1$. First,calculate $i^{22}$: $i^{22} = (i^4)^5 	imes i^2 = 1^5 	imes (-1) = -1$. Next,calculate $(\frac{1}{i})^{35}$: $\frac{1}{i} = -i$. So,$(-i)^{35} = -(i^{35}) = -(i^{32} 	imes i^3) = -(1 	imes -i) = i$. Now,substitute these values into the expression: $\{i^{22} - (\frac{1}{i})^{35}\}^2 = \{-1 - i\}^2$. Expand the square: $(-1 - i)^2 = (-1)^2 + (-i)^2 + 2(-1)(-i) = 1 + i^2 + 2i$. Since $i^2 = -1$,we get $1 - 1 + 2i = 2i$.

The value of $\{i^{22}-(\frac{1}{i})^{35}\}^2$ is

Similar Questions

The value of $\frac{i^{592}+i^{590}+i^{588}+i^{586}+i^{584}}{i^{582}+i^{580}+i^{578}+i^{576}+i^{574}}-1=$

If $n$ is a positive integer,then $\left( \frac{1 + i}{1 - i} \right)^{4n + 1} = $

The least positive integer $n$ for which $(1+i)^n=(1-i)^n$ is

$A$ value of $\theta$ for which $\frac{2 + 3i \sin \theta}{1 - 2i \sin \theta}$ is purely imaginary,is:

The real value of $\alpha$ for which $\frac{1-i \sin \alpha}{1+2 i \sin \alpha}$ is purely real is

Vedclass Test Series

Exam Paper Generator

Online Exam Module