If $\frac{(a+i b)^2}{a-i b}-\frac{(a-i b)^2}{a+i b}=x+i y$, then $x$ is equal to

$\frac{-2 b^3}{\left(a^2+b^2\right)^2}$

$\frac{6 a^2 b}{\left(a^2+b^2\right)^2}$

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

\frac{(a+i b)^2}{a-i b}-\frac{(a-i b)^2}{a+i b}=x+i y

, then

x

is equal to

View answer and explanation

Correct option: C

The polar form of the complex number

1+\sqrt{3} i

2\left(\cos \frac{{\pi}}{{3}}+i \sin \frac{{\pi}}{{3}}\right)

because the modulus is 2 and the argument is

\frac{{\pi}}{{3}}

Practice All Chapter MCQs →

{\left| {1 - i} \right|^x} = {2^x}

|z|=1

z \neq \pm 1

z, \frac{z}{1-z^2}

\left|z_r-r\right| \leq r, \forall r=1,2,3, \ldots, n

\left|\sum\limits_{r=1}^n z_r\right|

(z+a b)^3=a^3

a \neq 0

\alpha, \beta

\gamma

|\alpha-\beta|

{z^2} + {\left| z \right|^2} = 0,

z \in C

2x = - 1 + \sqrt 3 i

{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6}