Single Choice

Consider the complex number $$z_1$$ and $$z_2$$ satisfying the relation $$|z_1+z_2|^2=|z_1|^2+|z_2|^2$$, then one of the possible argument of complex number $$i\dfrac{z_1}{z_2}$$ is,

A$$\dfrac {\pi}{2}$$
B$$-\dfrac {\pi}{2}$$
C$$0$$
Correct Answer
Dnone of these

Solution

$$|z_{1}+z_{2}|^{2}=z_{1}^{2}+z_{2}^{2}$$

$$z_{1}^{2}+z_{2}^{2}+z_{1}\overline{z_{2}}+\overline{z_{1}}z_{2}=z_{1}^{2}+z_{2}^{2}$$

$$z_{1}\overline{z_{2}}+\overline{z_{1}}z_{2}=0$$

let
$$z_{1}=e^{i\theta}$$

$$z_{2}=e^{i\gamma}$$

Substituting, in the above equation, we get

$$e^{i(\theta-\gamma)}=-e^{-i(\theta-\gamma)}$$

$$e^{2i(\theta-\gamma)}=-1$$

Hence
$$2(\theta-\gamma)=(2n-1)\pi$$

$$\theta-\gamma=\dfrac{(2n-1)\pi}{2}$$

Now $$\theta-\gamma=arg(z_{1}\overline{z_{2}})$$

Therefore, $$arg(z_{1}\overline{z_{2}})=\dfrac{(2n-1)\pi}{2}$$

Hence $$z_{1}\overline{z_{2}}$$ is purely imaginary.

Therefore
$$i(z_{1}\overline{z_{2}})$$ will be purely real.

$$arg(i(z_{1}\overline{z_{2}}))$$ will be 0 or $$\pi$$.

Hence, option $$'C'$$ is correct.

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