Find the modulus and argument of the complex number $\frac{1+2 i}{1-3 i}$

Let $z=\frac{1+3 i}{1-3 i},$ then

$z=\frac{1+2 i}{1-3 i} \times \frac{1+3 i}{1+3 i}=\frac{1+3 i+2 i+6 i^{2}}{1^{2}+3^{2}}=\frac{1+5 i+6(-1)}{1+9}$

$=\frac{-5+5 i}{10}=\frac{-5}{10}+\frac{5 i}{10}=\frac{-1}{2}+\frac{1}{2} i$

Let $z=r \cos \theta+i r \sin \theta$

i.e., $r \cos \theta=\frac{-1}{2}$ and $r \sin \theta=\frac{1}{2}$

On squaring and adding, we obtain

$r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=\left(\frac{-1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}$

$\Rightarrow r^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$

$\Rightarrow r=\frac{1}{\sqrt{2}}$    $[\text { Conventionally, } r>0]$

$\therefore \frac{1}{\sqrt{2}} \cos \theta=\frac{-1}{2}$ and $\frac{1}{\sqrt{2}} \sin \theta=\frac{1}{2}$

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

$\therefore \theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4}$      [As $\theta$ lies in the $II$ quadrant]

Therefore, the modulus and argument of the given complex number are $\frac{1}{\sqrt{2}}$ and $\frac{3 \pi}{4}$ respectively.