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?
 A: 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.
A: So as I touched on in the comments, we begin by "rationalizing the denominator" in a sense, in that we multiply each fraction's top and bottom by the conjugate of the denominator. Thus,
$$\frac{3+4i}{1-i} + \frac{2-i}{2+3i} = \frac{3+4i}{1-i} \left( \frac{1+i}{1+i} \right) + \frac{2-i}{2+3i} \left( \frac{2-3i}{2-3i} \right) \tag 1$$
The bottoms simplify nicely since $(a-b)(a+b)=a^2 - b^2$. Alternatively, since we have complex numbers, $z \bar z = |z|^2$; whichever you prefer to use.
Anyhow, $(1)$ becomes, with simplification,
$$\frac{3+4i}{1-i} \left( \frac{1+i}{1+i} \right) + \frac{2-i}{2+3i} \left( \frac{2-3i}{2-3i} \right) = \frac{-1+7i}{2} + \frac{1-8i}{13} \tag 2$$
We combine the real and imaginary parts to get
$$-\frac{11}{26} + \frac{75}{26}i$$

Up to this point, you've done everything correctly.

The number is a little messy but it can be worked with. Let $a = -11/26, b = 75/26$. Then the above number is $a+bi$. We know that for numbers in this form that
$$|a+bi| = \sqrt{a^2 + b^2} \;\;\;\;\;\; \arg(a+bi) = atan2(b,a)$$
Both of these are fairly evident if you imagine plotting $a+bi$ on the plane. Then its magnitude is the distance from the origin, which you can easily find from the Pythagorean theorem, and its argument is the angle it makes with the positive real axis (with counterclockwise rotation being positive). The $atan2$ function can calculate this directly if you like, but the more "intuitive" definition (angle from the positive real axis) may be easier to contend with for you, all depending - you don't want to memorize that complex formula after all. You will have to use a little trig, regardless.
A rough image for a generic $z$ in the complex plane, to help further what I mean by these intuitive notions:

In any event, substitute $a,b$ into the expressions above and do the arithmetic to find the magnitude and argument.
If it were me finding the argument, I'd note that, since $a<0,b>0$, then $z=a+bi$ is in the top-left quadrant of the complex plane. Then I would find the angle $z$ makes with the imaginary axis and add $\pi/2$ radians to it. This diagram makes it clear why:

