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 ?

  • 5
    $\begingroup$ 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

Beside the actual answer (by Peter Foreman in comment), a hint: you don't really need to multiply $z_2\bar z_1$, as $\arg \bar w = - \arg w$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.