Question

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

• A. $ta{n}^{-1}\left(2\right)$
• B. $-ta{n}^{-1}\left(2\right)$
• C. $ta{n}^{-1}\left(\frac{1}{2}\right)$
  
• D. $-ta{n}^{-1}\left(\frac{1}{2}\right)$

Answer 1

The correct option is B

$-ta{n}^{-1}\left(2\right)$

$\because \frac{2+i}{4i+(1+i{)}^2}=\frac{2+i}{4i+1+i^2+2i}=\frac{2+i}{4i+1-1+2i}=\frac{2+i}{6i}$

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

$\theta =ta{n}^{-1}\frac{-\frac{1}{3}}{\frac{1}{6}}=ta{n}^{-1}(-2)=-ta{n}^{-1}\left(2\right)$

$\because$ our complex number lies in the fourth quadrant; principal argument = $-ta{n}^{-1}\left(2\right)$