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(B) Let $z = x + iy$. Then $\bar{z} = x - iy$ and $|z| = \sqrt{x^2 + y^2}$.Given equation: $(x - iy)^2 + \sqrt{x^2 + y^2} = 0$.$(x^2 - y^2 - 2ixy) + \

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $z = x + iy$. Then $\bar{z} = x - iy$ and $|z| = \sqrt{x^2 + y^2}$.Given equation: $(x - iy)^2 + \sqrt{x^2 + y^2} = 0$.$(x^2 - y^2 - 2ixy) + \sqrt{x^2 + y^2} = 0$.Equating the imaginary part to zero: $-2xy = 0 \implies x = 0$ or $y = 0$.Case $1$: If $x = 0$,then $-y^2 + |y| = 0 \implies |y|^2 = |y|$. Since $z 
eq 0$,$|y| = 1$,so $y = 1$ or $y = -1$. Thus,$z_1 = i$ and $z_2 = -i$.Case $2$: If $y = 0$,then $x^2 + |x| = 0 \implies |x|^2 + |x| = 0$. Since $|x| \geq 0$,this implies $|x| = 0$,so $x = 0$,which gives $z = 0$ (not a non-zero solution).The non-zero solutions are $z_1 = i$ and $z_2 = -i$.Sum $\alpha = i + (-i) = 0$.Product $\beta = i 	imes (-i) = -i^2 = 1$.Therefore,$4(\alpha^2 + \beta^2) = 4(0^2 + 1^2) = 4(1) = 4$.

Let $\alpha$ and $\beta$ be the sum and the product of all the non-zero solutions of the equation $(\bar{z})^2+|z|=0$,where $z \in \mathbb{C}$. Then $4(\alpha^2+\beta^2)$ is equal to:

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