JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Application +4 | -1

(1+z)^n=a_0+a_1 z+a_2 z^2+\cdots+a_n z^n

, where

a_0, a_1

a_2, \ldots, a_n

, are real, then

\left(a_0-a_2+a_4-a_6+\cdots\right)^2+\left(a_1-a_3+a_5-a_7+\cdots\right)^2=

View answer and explanation

Correct option: A

2^n

Option A (

2^n

) is correct because substituting z=i into the binomial expansion and comparing magnitudes leads to this result.

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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}