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

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(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

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$2$

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(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

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