Complex numbers argument problem What would be the argument of $z$ if $z=2016-2016i$? 
$i$ is iota. 
My try. The argument of $2016$ should be $zero$ as it lies on the positive x axis of argand plane. And the argument for $2016i$ should be $0$-$pi/2$=$pi/2$ since it lies on positive y axis of argand plane. Finally argument of $z$ should be -$pi/2$.
But the answer shown is -$pi/4$.Just the misprint or am I missing out on a concept? Would love to learn any concept related to it. 
 A: Arguments do not add together! Where in the world did you get the idea that $$\arg(z)+\arg(w)=\arg(z+w)?!$$ If this were true, every complex number would have an argument of either $0$, $\pi/2$ or $-\pi/2$, since, for example, $1+4i$ should have an argument of $0$ (the argument of $1$) plus $\pi/2$ (argument of $4i$), which sums to $\pi/2$.
I strongly suggest you re-read the chapter on arguments!

The argument of a number is the angle between the real axis and the number, so you can (by drawing) quickly see that the tan of the argument is simply the imaginary part, divided by the real part.

OK, I was a bit quick saying "Arguments do not add together". In fact, they do, but not while adding numbers. They add when you multiply numbers, so the correct equation would be
$$\arg(z)+\arg(w)=\arg(z\cdot w)$$
A: The argument $\theta$ of a complex number $z=a+bi$ is given by $\theta=\arctan({\frac ba})$. However, as with trigonometric functions, this can have infinite values, two of which satisfy the domain $\theta \in (-\pi,\pi]$, so you need to make sure you're choosing the one which is appropriate for the quadrant the complex number is in.
In this case, $$\arg{(z)}=\arctan({\frac{-2016}{2016}})=\arctan(-1)=-\frac{\pi}{4}$$ as $\theta$ is in the fourth quadrant.
A: Treat the terms 2016 and 2016i as vectors, now do you see it is something in the IInd quadrant at $45^\circ$ ? You can't add arguments like scalars.
