Question

Argument of complex number $\frac{\left(-1-3i\right)}{\left(2+i\right)}$ is

• A. ${45}^0$
• B. ${135}^0$
• C. ${225}^0$
• D. ${240}^0$

Answer 1

The correct option is C

${225}^0$

Explanation for the correct option:

Step 1: Simplify the given complex number

Let $Z=\frac{\left(-1-3i\right)}{\left(2+i\right)}$

$=\frac{-1-3i}{2+i}\times \frac{2-i}{2-i}$

$=\frac{-2-5i+3i^2}{4-i^2}$

$=-1-i$ ……..(1)

Step 2 : Find real and imaginary part of complex number

Let $Z=r\cos \theta +ir\sin \theta$ ………(2)

Compare real and imaginary part of equation (1) and (2)

$\Rightarrow r\cos \theta =-1,r\sin \theta =-1$

Step 3. square and add real and imaginary parts:

$r^2\left({\cos }^2\theta +{\sin }^2\theta \right)=2$

$\Rightarrow$ $r^2=2$

$\Rightarrow$ $r=\pm \sqrt{2}$

As, $r>0,r=\sqrt{2}$

Step 4. Put the value of $r$ in $r\cos \theta ,r\sin \theta$:

$\sqrt{2}\cos \theta =-1,\sqrt{2}\sin \theta =-1$

$\Rightarrow$ $\cos \theta =\frac{-1}{\sqrt{2}},\sin \theta =\frac{-1}{\sqrt{2}}$

Therefore, $\theta =225°$

Thus, argument of given complex number is $225°$

Hence, Option ‘C’ is Correct.