Question

Find the argument and the principal value of the argument of the complex number $z=\frac{2+i}{4i+{(1+i)}^2},$ where $i=\sqrt{-1}$

Answer 1

Since $z=\frac{2+i}{4i+{(1+i)}^2}$

$=\frac{2+i}{4i+1+i^2+2i}$

$=\frac{2+i}{4i+1-1+2i}$

$=\frac{2+i}{6i}$

$=\frac{1}{3i}\times \frac{i}{i}+\frac{1}{6}$

$=\frac{i}{-3}+\frac{1}{6}$

$=\frac{1}{6}-\frac{1}{3}i$

$\therefore z$ lies in $IV$ quadrant.

Here $\theta ={\tan }^{-1}\left|\frac{\frac{-1}{3}}{\frac{1}{6}}\right|={\tan }^{-1}2$

$\therefore$ arg$z=2\pi -\theta =2\pi -{\tan }^{-1}2$

Hence, principal value of arg$z=-\theta =-{\tan }^{-1}2$