If z is a complex number,then z cdot overline | vedclass

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(A) Let $z = x + iy$,where $x, y \in \mathbb{R}$. Then $\overline{z} = x - iy$.Given $z \cdot \overline{z} = 0$.Substituting the values,we get $(x + i

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$z = 0$

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(A) Let $z = x + iy$,where $x, y \in \mathbb{R}$. Then $\overline{z} = x - iy$.Given $z \cdot \overline{z} = 0$.Substituting the values,we get $(x + iy)(x - iy) = 0$.This simplifies to $x^2 - (iy)^2 = 0$,which is $x^2 + y^2 = 0$.Since $x^2 \ge 0$ and $y^2 \ge 0$ for all real $x$ and $y$,the sum $x^2 + y^2 = 0$ is possible if and only if $x = 0$ and $y = 0$.Therefore,$z = 0 + 0i = 0$.

If $z$ is a complex number,then $z \cdot \overline{z} = 0$ if and only if

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