If $\cos (1-i)=a+i b$, where $a, b \in R$ and $i=\sqrt{-1}$, then

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

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Application +4 | -1

\cos (1-i)=a+i b

, where

a, b \in R

and

i=\sqrt{-1}

, then

View answer and explanation

Correct option: B

a=\frac{1}{2}\left(e+\frac{1}{e}\right) \cos 1, b=\frac{1}{2}\left(e-\frac{1}{e}\right) \sin 1

Option B (

a=\frac{1}{2}\left(e+\frac{1}{e}\right) \cos 1, b=\frac{1}{2}\left(e-\frac{1}{e}\right) \sin 1

) is correct because it uses the correct hyperbolic trigonometric identities to expand

\cos(1-i)

and separate the real and imaginary parts.

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

|z|=1

z \neq \pm 1

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\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}