Finding the angle of $-3+5i$ To find the angle $\theta$ of this complex number I know that I have to imagine it in the complex graph, draw a triangle and then calculate the arctan.
Here is the representation of $-3+5i$ on the graph:

(I'll explain $\alpha$ and $\phi$ in a moment)
I know that since I have the adjacent and opposite sides of the triangle I can correlate them with the angle using the tangent. Then, I need to calculate the arctan to get the angle.
But my problem is finding the tangent, because I don't know which is the right triangle in this case.
My question is: 
 is $tan \theta = \frac{5}{-3}$ or $tan \theta = \frac{-3}{5}$?
How do I know whether if $\theta$ is supposed to be $\alpha$ or $\phi$?
 A: By convention your $\theta$ is defined to be 

the angle made with the positive $X$-axis in anticlockwise direction

In this case it will be $90^\circ+\alpha$
A: Angle of complex number is angle the complex vector makes with positive part of real (usually $x$) axis in counterslockwise direction. So, the answer is $\alpha+\frac\pi2$.
A: Remember that lengths are positive.
The green triangle with angle $\varphi$ has an adjacent of length $3$ and an opposite of length $5$.
Hence $\tan \varphi = \frac{5}{3}$ and so $\varphi = \arctan \frac{5}{3} \approx 1.03$.
The red triangle with angle $\alpha$ has an adjacent of length $5$ and an opposite of length $3$.
Hence $\tan \alpha = \frac{3}{5}$ and so $\alpha = \arctan \frac{3}{5} \approx 0.54$.
Notice that $\varphi + \alpha = \arctan \frac{5}{3} + \arctan \frac{3}{5} = \frac{1}{2}\pi$, as expected.
Now, the argument of a complex number is by definition the angle made with the positive real axis. So to find the argument of $-3+5\mathrm i$ you stand at the origin looking at the positive reals, e.g. at number $1 = 1+0\mathrm i$. To face the number $-3+5\mathrm i$ you need to turn anti-clockwise to the positive imaginary axis, and then continue on through the angle $\alpha$.
Hence, $\arg(-3+5\mathrm i) = \frac{1}{2}\pi + \alpha = \frac{1}{2}\pi+\arctan\frac{3}{5} \approx 2.11$ 
