If $ \left(\frac{1+i}{1-i}\right)^{x}=1 $ then

$ x=2 n+1 $, where $ \mathrm{n} $ is any positive integer

$ x=2 n $, where $ \mathrm{n} $ is any positive integer

$ x=4 n+1 $, where $ \mathrm{n} $ is any positive integer.

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

\left(\frac{1+i}{1-i}\right)^{x}=1

then

View answer and explanation

Correct option: B

x=4 n

, where

\mathrm{n}

is any positive integer

The correct answer is B (

x=4n

, where

n

is any positive integer) because it satisfies the condition

\left(\frac{1+i}{1-i}\right)^{x}=1

by simplifying to

i^x = 1

, which holds true for multiples of 4.

Practice All Chapter MCQs →

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z \neq \pm 1

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\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}