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If \(\arg \left(\bar{z}_1\right)=\arg \left(z_2\right)\), then
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Verified Answer
The correct answer is:
\(\mathrm{z}_2=\mathrm{kz}_1^{-1}(\mathrm{k} > 0)\)
\(\begin{aligned}
& \bar{z}_1=\frac{\mathrm{z}_1 \overline{\mathrm{z}}_1}{\mathrm{z}_1}=\left|\mathrm{z}_1\right|^2 \mathrm{z}_1^{-1} \\
& \Rightarrow \arg \left(\mathrm{z}_1^{-1}\right)=\arg \left(\overline{\mathrm{z}}_1\right)=\arg \left(\mathrm{z}_2\right) \\
& \Rightarrow \mathrm{z}_2=\mathrm{kz}_1^{-1}(\mathrm{k} > 0)
\end{aligned}\)
& \bar{z}_1=\frac{\mathrm{z}_1 \overline{\mathrm{z}}_1}{\mathrm{z}_1}=\left|\mathrm{z}_1\right|^2 \mathrm{z}_1^{-1} \\
& \Rightarrow \arg \left(\mathrm{z}_1^{-1}\right)=\arg \left(\overline{\mathrm{z}}_1\right)=\arg \left(\mathrm{z}_2\right) \\
& \Rightarrow \mathrm{z}_2=\mathrm{kz}_1^{-1}(\mathrm{k} > 0)
\end{aligned}\)
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