If |z_1 + z_2| = |z_1 - z_2|,then the differe | vedclass

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(C) Given $|z_1 + z_2| = |z_1 - z_2|$.Squaring both sides,we get $|z_1 + z_2|^2 = |z_1 - z_2|^2$.$(z_1 + z_2)(\overline{z_1} + \overline{z_2}) = (z_1

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$\frac{\pi}{2}$

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(C) Given $|z_1 + z_2| = |z_1 - z_2|$.Squaring both sides,we get $|z_1 + z_2|^2 = |z_1 - z_2|^2$.$(z_1 + z_2)(\overline{z_1} + \overline{z_2}) = (z_1 - z_2)(\overline{z_1} - \overline{z_2})$.$|z_1|^2 + z_1\overline{z_2} + \overline{z_1}z_2 + |z_2|^2 = |z_1|^2 - z_1\overline{z_2} - \overline{z_1}z_2 + |z_2|^2$.$2(z_1\overline{z_2} + \overline{z_1}z_2) = 0$,which implies $z_1\overline{z_2} + \overline{z_1}z_2 = 0$.This means $2Re(z_1\overline{z_2}) = 0$,so $z_1\overline{z_2}$ is purely imaginary.Let $z_1 = r_1 e^{i	heta_1}$ and $z_2 = r_2 e^{i	heta_2}$.Then $z_1\overline{z_2} = r_1 r_2 e^{i(	heta_1 - 	heta_2)} = r_1 r_2 (\cos(	heta_1 - 	heta_2) + i\sin(	heta_1 - 	heta_2))$.Since $z_1\overline{z_2}$ is purely imaginary,its real part must be zero: $\cos(	heta_1 - 	heta_2) = 0$.Therefore,$	heta_1 - 	heta_2 = \pm \frac{\pi}{2}$.

If $|z_1 + z_2| = |z_1 - z_2|$,then the difference in the amplitudes of $z_1$ and $z_2$ is

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