For any two complex numbers z_1, z_2,if |z_1 | vedclass

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(A) Given $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$. Using the property $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + 2|z_1||z_2| \cos(	heta_1 - 	heta_2)$,where $\

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$	ext{Re} \left( \frac{z_1}{z_2} ight) = 0$

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(A) Given $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$. Using the property $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2 + 2|z_1||z_2| \cos(	heta_1 - 	heta_2)$,where $	heta_1 = \arg(z_1)$ and $	heta_2 = \arg(z_2)$. Substituting this into the given equation: $|z_1|^2 + |z_2|^2 + 2|z_1||z_2| \cos(	heta_1 - 	heta_2) = |z_1|^2 + |z_2|^2$. This implies $2|z_1||z_2| \cos(	heta_1 - 	heta_2) = 0$. Assuming $z_1, z_2 
eq 0$,we get $\cos(	heta_1 - 	heta_2) = 0$. This means $	heta_1 - 	heta_2 = \pm \frac{\pi}{2}$. Therefore,$\arg\left( \frac{z_1}{z_2} ight) = \arg(z_1) - \arg(z_2) = \pm \frac{\pi}{2}$. Since the argument of $\frac{z_1}{z_2}$ is $\pm \frac{\pi}{2}$,the complex number $\frac{z_1}{z_2}$ is purely imaginary. Thus,$	ext{Re} \left( \frac{z_1}{z_2} ight) = 0$.

For any two complex numbers $z_1, z_2$,if $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$,then:

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