If $z = \sqrt 3 + i$, then the argument of ${z^2}{e^{z - i}}$ is equal to

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

z = \sqrt 3 + i

, then the argument of

{z^2}{e^{z - i}}

is equal to

View answer and explanation

Correct option: D

\pi /3

Option D (

\pi /3

) is correct because the argument of the complex number

{z^2}{e^{z - i}}

simplifies to

\frac{\pi}{3}

when

z = \sqrt{3} + i

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}