JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Application +4 | -1

The least positive integer

n

for which

{\left( {\frac{{1 + i}}{{1 - i}}} \right)^n} = \frac{2}{\pi }{\sin ^{ - 1}}\left( {\frac{{1 + {x^2}}}{{2x}}} \right),

where

x > 0,

View answer and explanation

Correct option: B

Option B (4) is correct because it is the smallest positive integer

n

that satisfies the equation after simplifying the complex number and the inverse sine function.

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{\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}