The imaginary part of frac{(1 + i)^2}{2 - i} | vedclass

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

Question

(C) We have $z = \frac{(1 + i)^2}{2 - i}$.Since $(1 + i)^2 = 1^2 + i^2 + 2i = 1 - 1 + 2i = 2i$, the expression becomes $z = \frac{2i}{2 - i}$.To simpl

Vedclass · Accepted Answer

$\frac{4}{5}$

Vedclass · Answer

(C) We have $z = \frac{(1 + i)^2}{2 - i}$.Since $(1 + i)^2 = 1^2 + i^2 + 2i = 1 - 1 + 2i = 2i$, the expression becomes $z = \frac{2i}{2 - i}$.To simplify, multiply the numerator and denominator by the conjugate of the denominator $(2 + i)$:$z = \frac{2i(2 + i)}{(2 - i)(2 + i)} = \frac{4i + 2i^2}{2^2 - i^2} = \frac{4i - 2}{4 + 1} = \frac{-2 + 4i}{5} = -\frac{2}{5} + i\frac{4}{5}$.Thus, the imaginary part $Im(z) = \frac{4}{5}$.

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

Similar Questions

If $\frac{2+3 i}{i-2}-\frac{4 i-3}{3+4 i}=x+i y$,then $3 x+y=$

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 ${\left( {\frac{{1 - i}}{{1 + i}}} \right)^{100}} = a + ib$,then

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module