A) \[-\pi \]
B) \[-\frac{\pi }{2}\]
C) \[\frac{\pi }{2}\]
D) 0
Correct Answer: D
Solution :
Let\[{{z}_{1}}={{r}_{1}}\]\[(\cos {{\theta }_{1}}+i\sin {{\theta }_{1}})\],\[{{z}_{2}}={{r}_{2}}\] \[(\cos {{\theta }_{2}}+i\sin {{\theta }_{2}})\] \[\therefore \]\[|{{z}_{1}}+{{z}_{2}}|=[{{({{r}_{1}}\cos {{\theta }_{1}}+{{r}_{2}}\cos {{\theta }_{2}})}^{2}}\]\[+{{({{r}_{2}}\sin {{\theta }_{1}}+{{r}_{2}}\sin {{\theta }_{2}})}^{2}}{{]}^{1/2}}\] \[={{[r_{1}^{2}+r_{2}^{2}+2{{r}_{1}}{{r}_{2}}\cos ({{\theta }_{1}}-{{\theta }_{2}})]}^{1/2}}={{[{{({{r}_{1}}+{{r}_{2}})}^{2}}]}^{1/2}}\]\[arg\,\,({{z}_{2}})=\theta \] Therefore \[\cos ({{\theta }_{1}}-{{\theta }_{2}})=1\Rightarrow {{\theta }_{1}}-{{\theta }_{2}}=0\Rightarrow {{\theta }_{1}}={{\theta }_{2}}\] Thus arg \[({{z}_{1}})-arg({{z}_{2}})=0\]. Trick: \[|{{z}_{1}}+{{z}_{2}}|\,=\,|{{z}_{1}}|+|{{z}_{2}}|\Rightarrow {{z}_{1}},{{z}_{2}}\]lies on same straight line. \[\therefore arg\,{{z}_{1}}=arg\,{{z}_{2}}\Rightarrow arg\,{{z}_{1}}-arg\,{{z}_{2}}=0\]
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