Finding the modulus and argument of complex numbers My problem:

Find the magnitude and argument of the following.
  $$\frac{3+4i}{1-i} + \frac{2-i}{2+3i}$$


My solution so far:
\begin{align}
\frac{3+4i}{1-i} + \frac{2-i}{2+3i} &= \frac{3+4i}{1-i} \times \frac{1+i}{1+i} + \frac{2-i}{2+3i} \times \frac{2-3i}{2-3i}\\
&=\frac{3+3i+4i+4(-1)}{1+1} + \frac{4-6i-2i+3(-1)}{4+9(-1)}\\
&= \frac{3+7i-4}{2} + \frac{4-8i-3}{13} \\
&= \frac{-1+7i}{2} + \frac{1-8i}{13}\\
&= \frac{-13+91i+2-16i}{26}
\end{align}
Thus the complex number is given by
$$ z = \frac{-11+75i}{26}=  \frac{-11}{26} + \frac {75}{26}i $$
Then, in order to find the modulus, I computed
$$|z| =\sqrt{x\cdot x + y\cdot y}$$
where $x = -11/26$ and $y = 75/26 $. Then,
$$|z| = \sqrt{\frac{-121}{676} + \frac {5625}{676}} = \sqrt{\frac{5746}{676}}$$

This is where I stopped.
How do I find the argument of $z$ from here?

in your solution, you find, correctly, $a$ and $b$ such that $\frac{3+4i}{1-i} + \frac{2-i}{2+3i} = a+bi$
but you need to find the modulus and the argument of the number. That is, you need to find $r>0$ and $\theta\in[0,2\pi)$ such that 
$$\frac{3+4i}{1-i} + \frac{2-i}{2+3i} = r(\cos \theta + i\sin \theta).$$
There is a simple way to converting between the standard $a+bi$ format and the latter polar format.
