Frac{{1 - 2i}}{{2 + i}} + frac{{4

English

Gujarati

Hindi

4-1.Complex numbers

easy

$\frac{{1 - 2i}}{{2 + i}} + \frac{{4 - i}}{{3 + 2i}} = $

$\frac{{24}}{{13}} + \frac{{10}}{{13}}i$

$\frac{{24}}{{13}} - \frac{{10}}{{13}}i$

$\frac{{10}}{{13}} + \frac{{24}}{{13}}i$

$\frac{{10}}{{13}} - \frac{{24}}{{13}}i$

Solution

(d) $\frac{{1 – 2i}}{{2 + i}} + \frac{{4 – i}}{{3 + 2i}} = \frac{{(1 – 2i)(3 + 2i) + (4 – i)(2 + i)}}{{(2 + i)(3 + 2i)}}$
$ = \frac{{50 – 120i}}{{65}} = \frac{{10}}{{13}} – \frac{{24}}{{13}}i$.

Standard 11

Mathematics

If ${z_1}$ and ${z_2}$ be two complex number, then $Re({z_1}{z_2}) = $

normal

Let the complex number $z=x+$ iy be such that $\frac{2 z-3 i}{2 z+i}$ is purely imaginary. If $x + y ^2=0$, then $y^4+y^2-y$ is equal to :

medium

(JEE MAIN-2023)

$\frac{{3 + 2i\sin \theta }}{{1 – 2i\sin \theta }}$ will be purely imaginary, if $\theta = $

[Where $n$ is an integer]

medium

(IIT-1976)

The smallest positive integer $n$for which ${(1 + i)^{2n}} = {(1 – i)^{2n}}$is

normal

Solve the equation $x^{2}+3 x+5=0$

medium

$\frac{{1 - 2i}}{{2 + i}} + \frac{{4 - i}}{{3 + 2i}} = $

Solution

Similar Questions

If ${z_1}$ and ${z_2}$ be two complex number, then $Re({z_1}{z_2}) = $

Let the complex number $z=x+$ iy be such that $\frac{2 z-3 i}{2 z+i}$ is purely imaginary. If $x + y ^2=0$, then $y^4+y^2-y$ is equal to :

$\frac{{3 + 2i\sin \theta }}{{1 – 2i\sin \theta }}$ will be purely imaginary, if $\theta = $ [Where $n$ is an integer]

The smallest positive integer $n$for which ${(1 + i)^{2n}} = {(1 – i)^{2n}}$is

Solve the equation $x^{2}+3 x+5=0$

Start a Free Trial Now

$\frac{{3 + 2i\sin \theta }}{{1 – 2i\sin \theta }}$ will be purely imaginary, if $\theta = $

[Where $n$ is an integer]