JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Application +4 | -1

\cos \left(i \log \frac{a-i b}{a+i b}\right)

is equal to

View answer and explanation

Correct option: B

\frac{a^2-b^2}{a^2+b^2}

Option B (

\frac{a^2-b^2}{a^2+b^2}

) is correct because it is the simplified form of

\cos(i \log \frac{a-ib}{a+ib})

using complex number properties.

More Questions from COMPLEX NUMBERS

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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}