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If $z$ and $w$ are complex numbers such that $\bar{z}-i \bar{w}=0$ and $\operatorname{Arg}(z w)=\frac{3 \pi}{4}$, then $\operatorname{Arg} z=$
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Verified Answer
The correct answer is:
$\frac{\pi}{8}$
We have,
$$
\begin{aligned}
\Rightarrow & i \bar{\omega} & =\bar{z} \Rightarrow \omega=\frac{1}{i} \bar{z} \\
\Rightarrow & \omega & =-\frac{1}{i} z \Rightarrow \omega=i z
\end{aligned}
$$
Now, we have
$$
\begin{aligned}
& \arg (z \omega)=\frac{3 \pi}{4} \\
& \Rightarrow \quad \arg (z(i z))=\frac{3 \pi}{4} \\
& \Rightarrow \quad \arg \left(i z^2\right)=\frac{3 \pi}{4} \\
& \Rightarrow \quad \arg (i)+\arg \left(z^2\right)=\frac{3 \pi}{4} \\
& {\left[\because \arg \left(z_1 z_2\right)=\arg \left(z_1\right)+\arg \left(z_2\right)\right]} \\
& \Rightarrow \quad \arg (i)+2 \arg (z)=\frac{3 \pi}{4} \quad\left[\because \arg \left(z^n\right)=n \arg (z)\right] \\
& \Rightarrow \quad \frac{\pi}{2}+2 \arg (z)=\frac{3 \pi}{4} \\
& \Rightarrow \quad \arg (z)=\frac{\pi}{8} \\
&
\end{aligned}
$$
$$
\begin{aligned}
\Rightarrow & i \bar{\omega} & =\bar{z} \Rightarrow \omega=\frac{1}{i} \bar{z} \\
\Rightarrow & \omega & =-\frac{1}{i} z \Rightarrow \omega=i z
\end{aligned}
$$
Now, we have
$$
\begin{aligned}
& \arg (z \omega)=\frac{3 \pi}{4} \\
& \Rightarrow \quad \arg (z(i z))=\frac{3 \pi}{4} \\
& \Rightarrow \quad \arg \left(i z^2\right)=\frac{3 \pi}{4} \\
& \Rightarrow \quad \arg (i)+\arg \left(z^2\right)=\frac{3 \pi}{4} \\
& {\left[\because \arg \left(z_1 z_2\right)=\arg \left(z_1\right)+\arg \left(z_2\right)\right]} \\
& \Rightarrow \quad \arg (i)+2 \arg (z)=\frac{3 \pi}{4} \quad\left[\because \arg \left(z^n\right)=n \arg (z)\right] \\
& \Rightarrow \quad \frac{\pi}{2}+2 \arg (z)=\frac{3 \pi}{4} \\
& \Rightarrow \quad \arg (z)=\frac{\pi}{8} \\
&
\end{aligned}
$$
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