If |z1 + z2|^2 = |z1|^2 + |z2|^2 , then z1/z2 is (1) purely real (2) purely imaginary(3) zero of purely imaginary (4) Neither real nor imaginary

Solution:

Given |z1 + z2|2 = |z1|2 + |z2|2

⇒ \((z_{1}+z_{2})+(\bar{z_{1}}+\bar{z_{2}})=\left | z_{1} \right |^{2}+\left | z_{2} \right |^{2}\)

⇒ \(\left | z_{1} \right |^{2}+\left | z_{2} \right |^{2}=\left | z_{1} \right |^{2}+\left | z_{2} \right |^{2}+z_{1}\bar{z_{2}}+z_{2}\bar{z_{1}}\)

⇒ \(0=z_{1}\bar{z_{2}}+z_{2}\bar{z_{1}}\)

⇒ \(z_{1}\bar{z_{2}}=-z_{2}\bar{z_{1}}\)

⇒ \(\frac{\bar{z_{1}}}{\bar{z_{2}}}=-\frac{z_{1}}{z_{2}}\)

⇒ \(\bar{(\frac{z_{1}}{z_{2}})}=-\frac{z_{1}}{z_{2}}\)

Thus z1/z2 is purely imaginary.

Hence option (2) is the answer.

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