If {z_1} and {z_2} are two non-zero complex n | vedclass

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(D) Given that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|$. This condition implies that the complex numbers ${z_1}$ and ${z_2}$ lie on the same ray originat

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(D) Given that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|$. This condition implies that the complex numbers ${z_1}$ and ${z_2}$ lie on the same ray originating from the origin in the complex plane. Let ${z_1} = {r_1}(\cos{	heta_1} + i\sin{	heta_1})$ and ${z_2} = {r_2}(\cos{	heta_2} + i\sin{	heta_2})$. Then $|{z_1} + {z_2}|^2 = (|{z_1}| + |{z_2}|)^2 = |{z_1}|^2 + |{z_2}|^2 + 2|{z_1}||{z_2}|$. Also,$|{z_1} + {z_2}|^2 = (z_1 + z_2)(\overline{z_1} + \overline{z_2}) = |z_1|^2 + |z_2|^2 + z_1\overline{z_2} + \overline{z_1}z_2 = |z_1|^2 + |z_2|^2 + 2	ext{Re}(z_1\overline{z_2})$. Comparing these,$2	ext{Re}(z_1\overline{z_2}) = 2|z_1||z_2|$,which means $\cos(	heta_1 - 	heta_2) = 1$. Thus,$	heta_1 - 	heta_2 = 0$,which implies $	ext{arg}({z_1}) - 	ext{arg}({z_2}) = 0$.

If ${z_1}$ and ${z_2}$ are two non-zero complex numbers such that $|{z_1} + {z_2}| = |{z_1}| + |{z_2}|,$ then $\text{arg}({z_1}) - \text{arg}({z_2})$ is equal to

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