Express the following in the form a+ib:i^{-35 | vedclass

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

Question

(A) Given expression is $i^{-35}$.We know that $i^{-35} = \frac{1}{i^{35}}$.Since $i^4 = 1$,we can write $i^{35} = i^{32} 	imes i^3 = (i^4)^8 	imes

Vedclass · Accepted Answer

$0+i$

Vedclass · Answer

(A) Given expression is $i^{-35}$.We know that $i^{-35} = \frac{1}{i^{35}}$.Since $i^4 = 1$,we can write $i^{35} = i^{32} 	imes i^3 = (i^4)^8 	imes i^3 = 1^8 	imes (-i) = -i$.Therefore,$i^{-35} = \frac{1}{-i}$.Multiplying the numerator and denominator by $i$,we get $\frac{1 	imes i}{-i 	imes i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i$.In the form $a+ib$,this is $0+1i$.

Express the following in the form $a+ib$:
$i^{-35}$

Similar Questions

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

The values of $x$ and $y$ satisfying the equation $\frac{(1 + i)x - 2i}{3 + i} + \frac{(2 - 3i)y + i}{3 - i} = i$ are

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

If $a+bi = \frac{i}{1-i}$,then $(a, b) =$

The value of the series $1 + i^2 + i^4 + i^6 + ..... + i^{2n}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Express the following in the form $a+ib$:$i^{-35}$