The sum of the squares of the modulus of all | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given equation: $\bar{z} = i z^{2} + z^{2} - z$Rearranging the terms: $z + \bar{z} = (1 + i) z^{2}$Let $z = x + iy$. Then $z + \bar{z} = 2x$.So,$2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given equation: $\bar{z} = i z^{2} + z^{2} - z$Rearranging the terms: $z + \bar{z} = (1 + i) z^{2}$Let $z = x + iy$. Then $z + \bar{z} = 2x$.So,$2x = (1 + i)(x + iy)^{2} = (1 + i)(x^{2} - y^{2} + 2xyi)$.$2x = (x^{2} - y^{2} - 2xy) + i(x^{2} - y^{2} + 2xy)$.Equating real and imaginary parts:$1) \ 2x = x^{2} - y^{2} - 2xy$$2) \ 0 = x^{2} - y^{2} + 2xy \Rightarrow x^{2} - y^{2} = -2xy$.Substitute $(2)$ into $(1)$: $2x = -2xy - 2xy = -4xy$.$2x(1 + 2y) = 0$,so $x = 0$ or $y = -1/2$.If $x = 0$,then from $(2)$,$y^{2} = 0 \Rightarrow y = 0$. Thus $z = 0$,$|z|^{2} = 0$.If $y = -1/2$,then from $(2)$,$x^{2} - (-1/2)^{2} = -2x(-1/2)$ $\Rightarrow x^{2} - 1/4 = x$ $\Rightarrow 4x^{2} - 4x - 1 = 0$.The roots are $x = \frac{4 \pm \sqrt{16 + 16}}{8} = \frac{4 \pm 4\sqrt{2}}{8} = \frac{1 \pm \sqrt{2}}{2}$.For these values,$|z|^{2} = x^{2} + y^{2} = x^{2} + 1/4$.From $4x^{2} - 4x - 1 = 0$,$x^{2} = x + 1/4$.So $|z|^{2} = x + 1/4 + 1/4 = x + 1/2$.For $x_{1} = \frac{1 + \sqrt{2}}{2}$,$|z_{1}|^{2} = \frac{1 + \sqrt{2}}{2} + \frac{1}{2} = \frac{2 + \sqrt{2}}{2} = 1 + \frac{\sqrt{2}}{2}$.For $x_{2} = \frac{1 - \sqrt{2}}{2}$,$|z_{2}|^{2} = \frac{1 - \sqrt{2}}{2} + \frac{1}{2} = \frac{2 - \sqrt{2}}{2} = 1 - \frac{\sqrt{2}}{2}$.Sum of squares of modulus: $0 + (1 + \frac{\sqrt{2}}{2}) + (1 - \frac{\sqrt{2}}{2}) = 2$.

The sum of the squares of the modulus of all complex numbers $z$ satisfying $\bar{z} = i z^{2} + z^{2} - z$ is equal to

Similar Questions

If $f(x)$ is a polynomial of degree $n$ with rational coefficients and $1+2i, 2-\sqrt{3}$ and $5$ are three roots of $f(x)=0$,then the least value of $n$ is

If $\cos A+\cos B+\cos C=0$ and $\sin A+\sin B+\sin C=0$,then $\cos (A-B)=$

If $3+i$ and $2-\sqrt{3}$ are the roots of the equation $f(x)=a_0+a_1 x+a_2 x^2+\ldots+a_{n} x^{n}$ where $a_0, a_1, \ldots, a_{n} \in \mathbb{Z}$,then the least value of $n$ and the value of $a_0$ are respectively:

If $P, Q$ and $R$ are angles of an isosceles triangle and $\angle P = \frac{\pi}{2}$, then the value of $\left(\cos \frac{P}{3} - i \sin \frac{P}{3}\right)^3 + (\cos Q + i \sin Q) (\cos R - i \sin R) + (\cos P - i \sin P) (\cos Q - i \sin Q) (\cos R - i \sin R)$ is:

If $2i$ is a root of $f(z) = z^4 + z^3 + 2z^2 + 4z - 8 = 0$,then which among the following cannot be a root of $f(z) = 0$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module