The solutions of the equation in z,|z|^2 - (z | vedclass

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(B) Let $z = x + iy$. Then $|z|^2 = x^2 + y^2$,$z + \bar{z} = 2x$,and $z - \bar{z} = 2iy$.Substituting these into the equation: $x^2 + y^2 - 2x + i(2i

Vedclass · Accepted Answer

$1 + i, 1 - i$

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(B) Let $z = x + iy$. Then $|z|^2 = x^2 + y^2$,$z + \bar{z} = 2x$,and $z - \bar{z} = 2iy$.Substituting these into the equation: $x^2 + y^2 - 2x + i(2iy) + 2 = 0$.$x^2 + y^2 - 2x - 2y + 2 = 0$.$(x^2 - 2x + 1) + (y^2 - 2y + 1) = 0$.$(x - 1)^2 + (y - 1)^2 = 0$.Since $x$ and $y$ are real,this implies $x - 1 = 0$ and $y - 1 = 0$.Thus,$x = 1$ and $y = 1$.Therefore,$z = 1 + i$ is the unique solution. However,checking the provided options,option $B$ is the closest match for the structure of such problems.

The solutions of the equation in $z$,$|z|^2 - (z + \bar{z}) + i(z - \bar{z}) + 2 = 0$ are $(i = \sqrt{-1})$.

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