JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Application +4 | -1

The complex number

\mathrm{z}=1+\mathrm{i}

is rotated through an angle

\frac{3 \pi}{2}

in anticlockwise direction about the origin and stretched by additional

\sqrt{2}

unit, then the new complex number is:

View answer and explanation

Correct option: D

2(1-i)

Option D (

2(1-i)

) is correct because it represents the complex number after rotation and stretching as described in the problem.

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

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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}