left(frac{1+i}{1-i}right)^4+left(frac{1-i}{1+ | vedclass

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

Question

(C) First,simplify the term $\frac{1+i}{1-i}$ by multiplying the numerator and denominator by the conjugate $(1+i)$:$\frac{1+i}{1-i} = \frac{(1+i)(1+i

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) First,simplify the term $\frac{1+i}{1-i}$ by multiplying the numerator and denominator by the conjugate $(1+i)$:$\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1+i^2+2i}{1-i^2} = \frac{1-1+2i}{1+1} = \frac{2i}{2} = i$Similarly,for the second term:$\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1+i^2-2i}{1-i^2} = \frac{1-1-2i}{1+1} = \frac{-2i}{2} = -i$Now,substitute these values into the original expression:$(i)^4 + (-i)^4 = i^4 + i^4 = 1 + 1 = 2$

$\left(\frac{1+i}{1-i}\right)^4+\left(\frac{1-i}{1+i}\right)^4$ is equal to

Similar Questions

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 roots of the equation $ix^2 - 4x - 4i = 0$ are

If $x, y \in R$ and $(x + iy)(3 + 2i) = 1 + i$,then $(x, y)$ is

The solution of the equation $|z| - z = 1 + 2i$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module