A) \[0\]
B) \[-\pi \]
C) \[-\frac{\pi }{2}\]
D) \[\frac{\pi }{2}\]
E) \[\pi \]
Correct Answer: A
Solution :
Given, \[|{{z}_{1}}+{{z}_{2}}|=|{{z}_{1}}|+|{{z}_{2}}|\] Let \[{{z}_{1}}={{x}_{1}}+i{{y}_{1}}\] \[{{z}_{2}}={{x}_{2}}+i{{y}_{2}}\] \[|({{x}_{1}}+{{x}_{2}})+i({{y}_{1}}+{{y}_{2}})|=|{{x}_{1}}+i{{y}_{1}}|\] \[+|{{x}_{2}}+i{{y}_{2}}|\] \[\sqrt{{{({{x}_{1}}+{{x}_{2}})}^{2}}+{{({{y}_{1}}+{{y}_{2}})}^{2}}}=\sqrt{x_{1}^{2}+y_{1}^{2}}\] \[+\sqrt{x_{2}^{2}+y_{2}^{2}}\] Squaring on both sides \[x_{1}^{2}+x_{2}^{2}+2{{x}_{1}}{{x}_{2}}+y_{1}^{2}+y_{2}^{2}+2{{y}_{1}}{{y}_{2}}\] \[=x_{1}^{2}+y_{1}^{2}+x_{2}^{2}+y_{2}^{2}+2\sqrt{(x_{1}^{2}+y_{1}^{2})(x_{2}^{2}+y_{2}^{2})}\] \[\Rightarrow \]\[2{{({{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}})}^{2}}=2\sqrt{(x_{1}^{2}+y_{1}^{2})(x_{2}^{2}+y_{2}^{2})}\] \[\Rightarrow \]\[{{({{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}})}^{2}}=(x_{1}^{2}+y_{1}^{2})(x_{2}^{2}+y_{2}^{2})\] \[\Rightarrow \]\[x_{1}^{2}x_{2}^{2}+y_{1}^{2}y_{2}^{2}+2{{x}_{1}}{{x}_{2}}{{y}_{1}}{{y}_{2}}\] \[=x_{1}^{2}x_{2}^{2}+x_{2}^{2}y_{1}^{2}+x_{1}^{2}y_{2}^{2}+y_{1}^{2}y_{2}^{2}\] \[\Rightarrow \] \[2{{x}_{1}}{{x}_{2}}{{y}_{1}}{{y}_{2}}=x_{2}^{2}y_{1}^{2}+x_{1}^{2}y_{2}^{2}\] \[\Rightarrow \] \[{{(-{{x}_{1}}{{y}_{1}}+{{y}_{1}}{{x}_{1}})}^{2}}=0\] ?? (i) \[=\frac{({{x}_{1}}+i{{y}_{1}})({{x}_{2}}-i{{y}_{2}})}{(x_{2}^{2}-y_{2}^{2})}\] \[=\frac{({{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}})}{(x_{2}^{2}-y_{2}^{2})}+i\frac{({{y}_{1}}{{x}_{2}}-{{x}_{1}}{{y}_{2}})}{(x_{2}^{2}-y_{2}^{2})}\] \[\arg \left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)={{\tan }^{-1}}\left( \frac{\operatorname{Im}({{z}_{1}}/{{z}_{2}})}{\operatorname{Re}({{z}_{1}}/{{z}_{2}})} \right)\] \[={{\tan }^{-1}}\left( \frac{{{y}_{1}}{{x}_{2}}-{{x}_{1}}{{y}_{2}}}{{{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}} \right)\] \[={{\tan }^{-1}}(0)=0\] [from Eq.(i)]
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