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(A) We are given $|z_1+z_2|^2 = |z_1|^2 + |z_2|^2$. Using the property $|z|^2 = z \bar{z}$, we can write: $(z_1+z_2)(\bar{z_1}+\bar{z_2}) = z_1 \bar{z

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$\operatorname{Re}\left(\frac{z_1}{z_2}\right)=0$

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(A) We are given $|z_1+z_2|^2 = |z_1|^2 + |z_2|^2$. Using the property $|z|^2 = z \bar{z}$, we can write: $(z_1+z_2)(\bar{z_1}+\bar{z_2}) = z_1 \bar{z_1} + z_2 \bar{z_2}$. Expanding the left side: $z_1 \bar{z_1} + z_1 \bar{z_2} + z_2 \bar{z_1} + z_2 \bar{z_2} = z_1 \bar{z_1} + z_2 \bar{z_2}$. Subtracting $|z_1|^2$ and $|z_2|^2$ from both sides, we get: $z_1 \bar{z_2} + z_2 \bar{z_1} = 0$. This can be written as $z_1 \bar{z_2} + \overline{z_1 \bar{z_2}} = 0$. Since $z + \bar{z} = 2 \operatorname{Re}(z)$, we have $2 \operatorname{Re}(z_1 \bar{z_2}) = 0$, which implies $\operatorname{Re}(z_1 \bar{z_2}) = 0$. Dividing by $|z_2|^2$, we get $\operatorname{Re}\left(\frac{z_1 \bar{z_2}}{z_2 \bar{z_2}}\right) = 0$, which simplifies to $\operatorname{Re}\left(\frac{z_1}{z_2}\right) = 0$.

For any two non-zero complex numbers $z_1$ and $z_2$, if $|z_1+z_2|^2=|z_1|^2+|z_2|^2$, then

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