The complex number z is given by z = (1+3i)/(p+qi), where p and q are real and p > q > 0. Given that Arg(z) = pi/4, show that p - 2q = 0. First method I used:
$$ z = \frac{\sqrt{10} cis (\arctan(3))}{\sqrt{p^2 + q^2} cis (\arctan (\frac qp))} $$
$$ \arctan 3 - \arctan \frac qp = \frac \pi4 $$
$$ - \arctan \frac qp = \frac \pi4 - \arctan 3 $$
$$ \arctan \frac qp = -\frac \pi4 + \arctan 3 $$
$$ \frac qp = \tan (-\frac \pi4 + \arctan 3) $$
$$ \frac qp = \frac 12 $$
$$ p = 2q $$
$$ p - 2q = 0 $$
Second method from book, which I then also used:
$$ z = \frac {(1+3i)(p-qi)}{(p+qi)(p-qi)} $$
$$ z = \frac {p + (3p-q)i + 3q}{p^2+q^2} $$
$ Arg(z) = \frac {\pi}{4} $ so $ z = |z| \cos (\frac {\pi}{4}) + i|z| \sin (\frac{\pi}{4}).$
Hence, Re(z) = Im(z)
$$ 1 = \frac {3p-q}{p+3q}$$
$$ p+3q=3p-q$$
$$ -2p = -4q$$
$$ p = 2q$$
$$p - 2q = 0$$
Here I have three questions: 
1. What is the significance of $p > q > 0$?  
2. Is my first method wrong or not preferred? 
3. For the second method, what if the given argument was so Re(z) ≠ Im(z)?
 A: $$\dfrac{1+3i}{p+iq}=\dfrac{(1+3i)(p-iq)}{p^2+q^2}=\dfrac{p+3q+i(3p-q)}{p^2+q^2}$$
Now using this, Arg$(z)=\arctan\dfrac{3p-q}{p+3q}$ if $p+3q>0$
$\arctan\dfrac{3p-q}{p+3q}=\dfrac\pi4\implies\dfrac{3p-q}{p+3q}=\tan\dfrac\pi4=1$
A: We can consider the locations of the numerator and denominator on the complex plane as vectors $ \ \langle \ 1, \ 3 \ \rangle \ $ and  $ \ \vec{w} \ = \ \langle \ p, \ q \ \rangle \ $ .  Since $ \ Arg \ z \ = \ \frac{\pi}{4} \ $ , this represents the included angle between the vectors ( $ \ \langle \ 1, \ 3 \ \rangle \ $ being counter-clockwise from $ \ \vec{w} \ $ ) , so we have the scalar ("dot") product
$$   \langle \ 1, \ 3 \ \rangle \ \cdot \ \vec{w} \ \ = \ \ \sqrt{10} \ | \ \vec{w} \ | \ \cos \frac{\pi}{4} $$
$$ \Rightarrow \ \ p \ + \ 3q \ \ = \ \ \sqrt{5} \ \sqrt{p^2 \ + \ q^2} $$ 
$$ \Rightarrow \  \ p^2 \ + \ 6 \ p \ q \ + \ 9 \ q^2 \ \ = \ \ 5 \ (p^2 \ + \ q^2 ) $$
$$ \Rightarrow \ \ 2 \ p^2 \ - \ 3 \ p \ q \ - \ 2 \ q^2 \ = \ 0 \ \ \Rightarrow \ \ (2 \ p \ + \ q) \ ( p \ - \ 2q ) \ = \ 0 \ \ . $$
We may reject the factor solution $ \ 2 \ p \ + \ q \ \ = \ \ 0 \ $ , as this put $ \ \vec{w} \ $ counter-clockwise from $ \langle \ 1, \ 3 \ \rangle \ $ (which would give us $ \ Arg \ z \ = \ - \frac{\pi}{4} \ $ ) .  This approach spares us the algebra of working with conjugate factors.
As to your first question, the imposed condition keeps both complex numbers in the first quadrant (and is what lets us toss out the $ \ 2 \ p \ + \ q \  $   solution.  Your first method looks to be correct. There is not any special reason to prefer any of the methods discussed here over others. Yours is a more "geometric" method than the one the book uses.
A: 
Given is:

*

*$$\text{z}=\frac{1+3i}{p+qi}$$

*$$p\space\wedge\space q\in\mathbb{R}$$

*$$p>q>0\to p>0,0<q<p$$

*$$\arg\left[\text{z}\right]=\arg\left[\frac{1+3i}{p+qi}\right]=\frac{\pi}{4}$$


Now, step-by-step:

*

*The number $\text{z}$:

$$\text{z}=\frac{1+3i}{p+qi}=\frac{(1+3i)(p-qi)}{(p+qi)(p-qi)}=\frac{(1+3i)(p-qi)}{\left|p+qi\right|^2}=\frac{(1+3i)(p-qi)}{p^2+q^2}=$$
$$\frac{p+3q+(3p-q)i}{p^2+q^2}=\frac{p+3q}{p^2+q^2}+\frac{(3p-q)i}{p^2+q^2}$$

*

*The argument of the number $\text{z}$:

$$\arg\left[\text{z}\right]=\frac{\pi}{4}\Longleftrightarrow\arg\left[\frac{1+3i}{p+qi}\right]=\frac{\pi}{4}\Longleftrightarrow$$
$$\arg\left[1+3i\right]-\arg\left[p+qi\right]=\frac{\pi}{4}\Longleftrightarrow\arctan(3)-\arg\left[p+qi\right]=\frac{\pi}{4}\Longleftrightarrow$$
$$-\arg\left[p+qi\right]=\frac{\pi}{4}-\arctan(3)\Longleftrightarrow\arg\left[p+qi\right]=\arctan(3)-\frac{\pi}{4}\Longleftrightarrow$$
$$\arctan\left(\frac{q}{p}\right)=\arctan(3)-\frac{\pi}{4}\Longleftrightarrow\tan\left(\arctan\left(\frac{q}{p}\right)\right)=\tan\left(\arctan(3)-\frac{\pi}{4}\right)\Longleftrightarrow$$
$$\frac{q}{p}=\frac{1}{2}\Longleftrightarrow p-2q=0$$
