Expand $(4+i)(5+3i)$ and show $\pi/4=\arctan{1/4}+\arctan{3/5}$ I can't remember ever having done this before so if someone could help me out that would be great. The question is expand $(4+i)(5+3i)$ and hence show that $\pi/4=\arctan{1/4}+\arctan{3/5}$. 
Expansion: $$(4+i)(5+3i)=20+17i-3=17(1+i)=17\sqrt{2}e^{i\frac{\pi}{4}}.$$
But I don't see how this shows me anything about $\arctan$. Any pointers would be appreciated. :)
 A: HINT:
We know $\arg(z_1z_2)=\arg(z_1)+\arg(z_2)$  if  $0\le \arg(z_1)+\arg(z_2)\le\frac\pi2$
Now, 
$\arg(17+17i)=\arctan\frac{17}{17}=\arctan1=\frac\pi4$ as the Principal value of $\arctan$ lies in $\in\left[-\frac\pi2,\frac\pi2\right]$
$\arg(4+i)=\arctan\frac14<\arctan1=\frac\pi4$ and $\arg(5+3i)=\arctan\frac35<\arctan1=\frac\pi4$

Alternatively, from Page#$276$ of this,
$$\arctan a+\arctan b=\arctan \left(\frac{a+b}{1-ab}\right)\text{  if }ab<1$$
Here $a=\frac14,b=\frac35\implies ab=\frac14\cdot\frac35<1$
A: If we write a complex number in its various forms$$a+bi=re^{i\theta}=r(\cos\theta+i\sin\theta)=r\cos\theta+ir\sin\theta$$ we see that $\tan\theta=\cfrac ba$ so that $\theta=\arctan\cfrac ba$
So then if $$(a+ib)(c+id)=r_1e^{i\theta_1}r_2e^{i\theta_2}=r_1r_2e^{i(\theta_1+\theta_2)}=e+if$$ we have $$\arctan \frac fe=\theta_1+\theta_2=\arctan \frac ba+\arctan\frac dc$$
We just need to take a little care to work compatibly with the angles and arctan function - this is not a problem if we remain in the first quadrant. In other cases we may end up working with a value other than the principal value.
