If |z_1+z_2|^2=|z_1|^2+|z_2|^2,where z_1 and | vedclass

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(B) We know that $|z_1+z_2|^2 = (z_1+z_2)(\overline{z_1}+\overline{z_2}) = z_1\overline{z_1} + z_1\overline{z_2} + z_2\overline{z_1} + z_2\overline{z_

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$\frac{z_1}{z_2}$ is purely imaginary

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(B) We know that $|z_1+z_2|^2 = (z_1+z_2)(\overline{z_1}+\overline{z_2}) = z_1\overline{z_1} + z_1\overline{z_2} + z_2\overline{z_1} + z_2\overline{z_2} = |z_1|^2 + |z_2|^2 + z_1\overline{z_2} + \overline{z_1\overline{z_2}}$.Given $|z_1+z_2|^2 = |z_1|^2 + |z_2|^2$,we have $z_1\overline{z_2} + \overline{z_1\overline{z_2}} = 0$.This implies $2 	ext{Re}(z_1\overline{z_2}) = 0$,so $z_1\overline{z_2}$ is purely imaginary.Let $z_1\overline{z_2} = ki$ for some $k \in \mathbb{R}$.Then $\frac{z_1}{z_2} = \frac{z_1\overline{z_2}}{|z_2|^2} = \frac{ki}{|z_2|^2}$,which is purely imaginary.

If $|z_1+z_2|^2=|z_1|^2+|z_2|^2$,where $z_1$ and $z_2$ are two complex numbers,then

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