Single Choice

If $$z_{1}$$ and $$z_{2}$$ be two complex numbers such that $$|z_{1}+z_{2}|=|z_{1}-z_{2}|$$ then amp $$(z_{1})-amp (z_{2})=?$$

A$$\dfrac{\pi}{4}$$
B$$\dfrac{\pi}{3}$$
C$$\dfrac{\pi}{2}$$
Correct Answer
D$$\pi$$

Solution

Let $$z_{1}=(x_{1}+iy_{1})$$ nad $$z_{2}=(x_{2}+iy_{2})$$. Then,$$(z_{1}+z_{2})=(x_{1}+x_{2})+i(y_{1}+y_{2})$$ and $$(z_{1}-z_{2})=(x_{1}-x_{2})+i(y_{1}-y_{2})$$Now, $$|z_{1}+z_{2}|=|z_{1}-z_{2}|$$\Rightarrow |z_{1}+z_{2}|^{2}=|z_{1}-z_{2}|^{2}\Rightarrow (x_{1}+x_{2})^{2}+(y_{1}+y_{2})^{2}=(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}$$\Rightarrow (x_{1}+x_{2})^{2}-(x_{1}-x_{2})^{2}+(y_{1}+y_{2})^{2}-(y_{1}-y_{2})^{2}=0$$\Rightarrow 4(x_{1}x_{2}+y_{1}y_{2})=0\Rightarrow (x_{1}x_{2}+y_{1}y_{2})=0$$\therefore amp (z_{1})-amp (z_{2})=\tan^{-1}\dfrac{y_{1}}{x_{1}}-\tan^{-1}\dfrac{y_{2}}{x_{2}}$$=\tan^{-1}\left\{\dfrac{\left(\dfrac{y_{1}}{x_{1}}-\dfrac{y_{2}}{x_{2}}\right)}{1+\left(\dfrac{y_{1}}{x_{1}}-\dfrac{y_{2}}{x_{2}}\right)}\right\}=\tan^{-1}\left\{\dfrac{(x_{2}y_{1}-x_{1}y_{2})}{(x_{1}x_{2}+y_{1}y_{2})}\right\}$$=\tan^{-1}\infty=\dfrac{\pi}{2}$$ $$\left[\because x_{1}x_{2}+y_{1}y_{2}=0\right]$$.

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