If $x + iy = \frac{3}{{2 + \cos {\rm{\theta }} + i\sin {\rm{\theta }}}},\;\,{\rm{then}}\,\;{x^2} + {y^2}$ is equal to

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

x + iy = \frac{3}{{2 + \cos \mathrm{\theta } + i\sin \mathrm{\theta }}},\;\,\mathrm{then}\,\;{x^2} + {y^2}

is equal to

View answer and explanation

Correct option: B

4\;x - 3

Option B (

4x - 3

) is correct because after rationalizing the denominator and equating real and imaginary parts,

{x^2} + {y^2}

simplifies to

4x - 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}