The conjugate of frac{(2 + i)^2}{3 + i} in th | vedclass

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

Question

(C) Let $z = \frac{(2 + i)^2}{3 + i}$.First,expand the numerator: $(2 + i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i$.So,$z = \frac{3 + 4i}{3 + i}$.To si

Vedclass · Accepted Answer

$\frac{13}{10} + i\left(-\frac{9}{10}\right)$

Vedclass · Answer

(C) Let $z = \frac{(2 + i)^2}{3 + i}$.First,expand the numerator: $(2 + i)^2 = 4 + 4i + i^2 = 4 + 4i - 1 = 3 + 4i$.So,$z = \frac{3 + 4i}{3 + i}$.To simplify,multiply the numerator and denominator by the conjugate of the denominator $(3 - i)$:$z = \frac{(3 + 4i)(3 - i)}{(3 + i)(3 - i)} = \frac{9 - 3i + 12i - 4i^2}{3^2 + 1^2} = \frac{9 + 9i + 4}{9 + 1} = \frac{13 + 9i}{10} = \frac{13}{10} + i\frac{9}{10}$.The conjugate of $z = a + ib$ is $\bar{z} = a - ib$.Therefore,the conjugate is $\frac{13}{10} - i\frac{9}{10}$.

The conjugate of $\frac{(2 + i)^2}{3 + i}$ in the form of $a + ib$ is

Similar Questions

If the conjugate of $(x + iy)(1 - 2i)$ is $1 + i$,then

If ${z_1}, {z_2}, {z_3}$ are complex numbers such that $|{z_1}| = |{z_2}| = |{z_3}| = \left| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \right| = 1$,then $|{z_1} + {z_2} + {z_3}|$ is

Let ${z_1}$ be a complex number with $|{z_1}| = 1$ and ${z_2}$ be any complex number,then $\left| \frac{{z_1 - z_2}}{{1 - z_1 \bar{z}_2}} \right| = $

The real part of $(1 - \cos \theta + 2i\sin \theta )^{-1}$ is

The conjugate of the complex number $\frac{2 + 5i}{4 - 3i}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module