Question

If $z=\frac{-2(1+2i)}{3+i}$ where $i=\sqrt{-1}$, then the argument $\theta (-\pi <\theta \le \pi )$ of $z$ is

• A. $\frac{3\pi }{4}$
• B. $\frac{\pi }{4}$
• C. $\frac{5\pi }{6}$
• D. $-\frac{3\pi }{4}$

Answer 1

Answer: (D)

$-\frac{3\pi }{4}$

$z=\frac{-2(1+2i)}{3+i}$

Multipyling and dividing the equation with the conjugate of $(3+i)$ i.e $(3-i)$, we get

$z=\frac{-2(1+2i)}{3+i}=\frac{-2(1+2i)(3-i)}{(3+i)(3-i)}=\frac{-2(5+5i)}{10}=-1-i$

Argument $\theta ={tan}^{-1}\left(\frac{-1}{-1}\right)$

As both the real as well as the imaginary part are negative the argument lies in the third quadrant

$\therefore$ $\theta =\frac{-3\pi }{4}$