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

(N/A) Let $z = \frac{1+2i}{1-3i}$.
To simplify,multiply the numerator and denominator by the conjugate of the denominator,$1+3i$:
$z = \frac{1+2i}{1-3i} \times \frac{1+3i}{1+3i} = \frac{1 + 3i + 2i + 6i^2}{1^2 + 3^2} = \frac{1 + 5i - 6}{1 + 9} = \frac{-5 + 5i}{10} = -\frac{1}{2} + \frac{1}{2}i$.
Let $z = r(\cos \theta + i \sin \theta)$.
Then $r = |z| = \sqrt{(-\frac{1}{2})^2 + (\frac{1}{2})^2} = \sqrt{\frac{1}{4} + \frac{1}{4}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
For the argument $\theta$,we have $\cos \theta = \frac{-1/2}{1/\sqrt{2}} = -\frac{1}{\sqrt{2}}$ and $\sin \theta = \frac{1/2}{1/\sqrt{2}} = \frac{1}{\sqrt{2}}$.
Since $\cos \theta < 0$ and $\sin \theta > 0$,$\theta$ lies in the $II$ quadrant.
Thus,$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
The modulus is $\frac{1}{\sqrt{2}}$ and the argument is $\frac{3\pi}{4}$.