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JEE MATHEMATICS MCQ: COMPLEX NUMBERS

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Analyzing +4 | -1

If

b+i c=(1+a) z

and

a^2+b^2+c^2=1

, then

\frac{1+i z}{1-i z}

equals

View answer and explanation

Correct option: D

\frac{a+i b}{1+c}

Option D (

\frac{{a+i b}}{{1+c}}

) is correct because, after substituting

z = \frac{{b+ic}}{{1+a}}

and simplifying, the expression

\frac{{1+iz}}{{1-iz}}

reduces to

\frac{{a+ib}}{{1+c}}

.

More Questions from COMPLEX NUMBERS

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The number of non-zero integer solutions of the equation

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

|z|=1

z \neq \pm 1

, then for any complex number

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|

is less than The roots of the cubic equation

(z+a b)^3=a^3

a \neq 0

represents the vertices of a triangle. If

\alpha, \beta

\gamma

are the roots, then find the value of

|\alpha-\beta|

. Number of solutions of the equation

{z^2} + {\left| z \right|^2} = 0,

z \in C

2x = - 1 + \sqrt 3 i

, then the value of

{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6}