If z is a complex number such that z^2 = (bar | vedclass

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(C) Let $z = x + iy$,where $x, y \in \mathbb{R}$. Then its conjugate is $\bar{z} = x - iy$.Given the equation $z^2 = (\bar{z})^2$.Substituting the val

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Either $z$ is purely real or purely imaginary

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(C) Let $z = x + iy$,where $x, y \in \mathbb{R}$. Then its conjugate is $\bar{z} = x - iy$.Given the equation $z^2 = (\bar{z})^2$.Substituting the values: $(x + iy)^2 = (x - iy)^2$.Expanding both sides: $x^2 - y^2 + 2ixy = x^2 - y^2 - 2ixy$.Subtracting $x^2 - y^2$ from both sides: $2ixy = -2ixy$.Adding $2ixy$ to both sides: $4ixy = 0$.This implies $xy = 0$.Therefore,either $x = 0$ (which means $z$ is purely imaginary) or $y = 0$ (which means $z$ is purely real).Thus,$z$ is either purely real or purely imaginary.

If $z$ is a complex number such that $z^2 = (\bar{z})^2$,then

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