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If $z=\dfrac{(z_1+\bar{z}_2)z_1}{z_2\bar{z}_1}$ where $z_1=1+2i$ and $z_2=1-i$, then find $\arg(z)$

My Attempt $$ z_1+\bar{z}_2=2+3i,\quad(z_1+\bar{z}_2)z_1=(2+3i)(1+2i)=-4+7i\\ z_2\bar{z}_1=(1-i)(1-2i)=-1-3i\\ \arg(z)=\arg(z_1+\bar{z}_2)+\arg(z_1)-\arg(z_2\bar{z}_1)\\ =\Big[\tan^{-1}\frac{3}{2}+\tan^{-1}\frac{2}{1}\Big]-\tan^{-1}\frac{3}{1}\\ =\pi+\tan^{-1}\frac{7}{-4}-\tan^{-1}3=\pi-\Big[\tan^{-1}\frac{7}{4}+\tan^{-1}3\Big]\\ =\tan^{-1}\frac{19}{17} $$ OR $$ z=\dfrac{(z_1+\bar{z}_2)z_1}{z_2\bar{z}_1}=\frac{-4+7i}{-1-3i}.\frac{-1+3i}{-1+3i}=\frac{-17-19i}{10}\\ \arg(z)=\tan^{-1}\frac{19}{17} $$ My reference gives the solution $\arg(z)=-\pi+\tan^{-1}\frac{19}{17}$, what am I missing here ?

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    $\begingroup$ Yes the argument is arctan(19/17) but not the principal value as the result is in the 3rd quadrant. So we must take away Pi to get the actual value of the argument. $\endgroup$ – Peter Foreman Feb 4 at 21:50
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Beside the actual answer (by Peter Foreman in comment), a hint: you don't really need to multiply $z_2\bar z_1$, as $\arg \bar w = - \arg w$.

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