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Let \(z_1, z_2\) be two complex numbers such that \(\bar{z}_1-i \bar{z}_2=0\) and \(\arg \left(z_1 z_2\right)=\frac{3 \pi}{4}\), then \(\arg \left(z_1\right)=\)
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Verified Answer
The correct answer is:
\(\frac{\pi}{8}\)
Given \(\bar{z}_1-i \bar{z}_2=0\)
\(\begin{array}{ll}
\Rightarrow & \bar{z}_1=i \bar{z}_2 \\
\Rightarrow & \bar{z}_1=\overline{i z_2} \\
\Rightarrow & z_1=-i z_2
\end{array}\)
Clearly argument of \(z_1=\) argument of \(z_2-\frac{\pi}{2}\) or argument \(z_1=\) argument \(z_2-\frac{\pi}{2}\)
\(\Rightarrow\) argument \(z_2=\operatorname{argument} z_1+\frac{\pi}{2}\)
Let argument \(z_1=\alpha\)
Then, given argument \(\left(z_1 z_2\right)=\frac{3 \pi}{4}\)
\(\Rightarrow\) argument \(z_1+\operatorname{argument} z_2=\frac{3 \pi}{4}\)
\(\begin{aligned}
\alpha+\alpha+\frac{\pi}{2} & =\frac{3 \pi}{4} \\
2 \alpha & =\frac{3 \pi}{4}-\frac{\pi}{2} \Rightarrow \alpha=\frac{\pi}{8}
\end{aligned}\)
\(\begin{array}{ll}
\Rightarrow & \bar{z}_1=i \bar{z}_2 \\
\Rightarrow & \bar{z}_1=\overline{i z_2} \\
\Rightarrow & z_1=-i z_2
\end{array}\)
Clearly argument of \(z_1=\) argument of \(z_2-\frac{\pi}{2}\) or argument \(z_1=\) argument \(z_2-\frac{\pi}{2}\)
\(\Rightarrow\) argument \(z_2=\operatorname{argument} z_1+\frac{\pi}{2}\)
Let argument \(z_1=\alpha\)
Then, given argument \(\left(z_1 z_2\right)=\frac{3 \pi}{4}\)
\(\Rightarrow\) argument \(z_1+\operatorname{argument} z_2=\frac{3 \pi}{4}\)
\(\begin{aligned}
\alpha+\alpha+\frac{\pi}{2} & =\frac{3 \pi}{4} \\
2 \alpha & =\frac{3 \pi}{4}-\frac{\pi}{2} \Rightarrow \alpha=\frac{\pi}{8}
\end{aligned}\)
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