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If $\mathrm{z}_1, \mathrm{z}_2$ and $\mathrm{z}_3, \mathrm{z}_4$ are 2 pairs of complex conjugate numbers, then $\arg \left(\frac{z_1}{z_4}\right)+\arg \left(\frac{z_2}{z_3}\right)$ equals:
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Consider $\arg \left(\frac{z_1}{z_4}\right)+\arg \left(\frac{z_2}{z_3}\right)$
$$
=\arg \left(z_1\right)-\arg \left(z_4\right)+\arg \left(z_2\right)-\arg \left(z_3\right)
$$
$$
\begin{aligned}
&=\left(\arg \left(z_1\right)+\arg \left(z_2\right)\right)-\left(\arg \left(z_3\right)+\arg \left(z_4\right)\right) \\
&\quad \operatorname{given}\left(\begin{array}{l}
z_2=\bar{z}_1 \ \\
z_4=\bar{z}_3
\end{array}\right) \\
&=\left(\arg \left(z_1\right)+\arg \left(\bar{z}_1\right)\right)-\left(\arg \left(z_3\right)+\arg \left(\bar{z}_3\right)\right) \\
&\qquad\left\{\begin{array}{l}
\operatorname{also}\left(\arg \left(\bar{z}_1\right)=-\arg \left(z_1\right)\right. \\
\arg \left(\bar{z}_3\right)=-\arg \left(z_3\right)
\end{array}\right\} \\
&=\left(\arg \left(z_1\right)-\arg \left(z_1\right)\right)-\left(\arg \left(z_3\right)-\arg \left(z_3\right)\right) \\
&=0-0=0
\end{aligned}
$$
$$
=\arg \left(z_1\right)-\arg \left(z_4\right)+\arg \left(z_2\right)-\arg \left(z_3\right)
$$
$$
\begin{aligned}
&=\left(\arg \left(z_1\right)+\arg \left(z_2\right)\right)-\left(\arg \left(z_3\right)+\arg \left(z_4\right)\right) \\
&\quad \operatorname{given}\left(\begin{array}{l}
z_2=\bar{z}_1 \ \\
z_4=\bar{z}_3
\end{array}\right) \\
&=\left(\arg \left(z_1\right)+\arg \left(\bar{z}_1\right)\right)-\left(\arg \left(z_3\right)+\arg \left(\bar{z}_3\right)\right) \\
&\qquad\left\{\begin{array}{l}
\operatorname{also}\left(\arg \left(\bar{z}_1\right)=-\arg \left(z_1\right)\right. \\
\arg \left(\bar{z}_3\right)=-\arg \left(z_3\right)
\end{array}\right\} \\
&=\left(\arg \left(z_1\right)-\arg \left(z_1\right)\right)-\left(\arg \left(z_3\right)-\arg \left(z_3\right)\right) \\
&=0-0=0
\end{aligned}
$$
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