If $(4+i)(z+\bar{z})-(3+i)(z-\bar{z})+26 i=$ 0 , then the value of $|z|^2$ is

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Application +4 | -1

(4+i)(z+\bar{z})-(3+i)(z-\bar{z})+26 i=

0 , then the value of

|z|^2

View answer and explanation

Correct option: B

Option B (17) is correct because by substituting

z = x + iy

into the given equation and solving the resulting system of equations, we find

|z|^2 = x^2 + y^2 = 17

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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}