We have , $\frac{1+2i}{1-3i}$ = $\frac{1+2i}{1-3i}$ × $\frac{1+3i}{1+3i}$ = $\frac{1+3i+2i-6}{1+9}$ = $\frac{-5+5i}{10}$ = - $\frac{1}{2} + \frac{1}{2} i$ Let - $\frac{1}{2} = r \cos \theta$ ... (i) and $\frac{1}{2}$ = r sin θ ... (ii) Squaring and adding (i) and (ii), we get $r^2(\cos^2 \theta + \sin^2 \theta)$ = $\frac{1}{4} + \frac{1}{4}$ = $\frac{2}{4}$ = $\frac{1}{2}$ ⇒ $r^2$ = $\frac{1}{2}$ ⇒ r = $\frac{1}{\sqrt{2}}$ Substituting the value of r in (i) and (ii), we get $\frac{1}{\sqrt{2}}$ cos θ = $-\frac{1}{2} , \frac{1}{\sqrt{2}}$ sin θ = $\frac{1}{2}$ ⇒ cos θ = $-\frac{1}{\sqrt{2}}$ , sin θ = $\frac{1}{\sqrt{2}}$ ⇒ cos θ = - cos $\left( \frac{\pi}{4} \right)$ , sin θ = sin $\frac{\pi}{4}$ Here, cos θ < 0 , sin θ > 0. ∴ θ lies in second quadrant. θ = π - $\frac{\pi}{4}$ = $\frac{3\pi}{4}$ ∴ Modulus is $\frac{1}{\sqrt{2}}$ and argument is $\frac{3\pi}{4}$