A) 0
B) Purely imaginary
C) Purely real
D) None of these
Correct Answer: A
Solution :
We have \[arg\left( \frac{{{z}_{1}}}{{{z}_{2}}} \right)=\pi \] Þ \[arg({{z}_{1}})-arg({{z}_{2}})=\pi \] Þ \[arg\,\,({{z}_{1}})=arg\,\,({{z}_{2}})+\pi \] Let \[arg\,\,({{z}_{2}})=\theta \], then \[arg\]\[({{z}_{1}})=\pi +\theta \] \ \[{{z}_{1}}=|{{z}_{1}}|[\cos (\pi +\theta )+i\sin (\pi +\theta )]\]\[=|{{z}_{1}}|(-\cos \theta -i\sin \theta )\] and \[{{z}_{2}}=|{{z}_{2}}|(\cos \theta +i\sin \theta )\]\[=|{{z}_{1}}|(\cos \theta +i\sin \theta )\] \[(\because |{{z}_{1}}|=|{{z}_{2}}|)\] Hence\[{{z}_{1}}+{{z}_{2}}=0\].
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