Find the argument of $\frac{-1 + \sqrt3 i}{2+2i}$

I rewrite equation $\frac{-1 + \sqrt3 i}{2+2i}$ as $$\frac{ \sqrt3 - 1}{4} + \frac{ \sqrt3 + 1}{4} i$$ using the conjugacy technique.

And set forward to find the argument of this complex function. I'm assuming the argument is the angle between the positive real-axis and the line?

I do this by trying to calculate: $$\mathbf{cos}^{-1} \frac{ \sqrt3 + 1}{\sqrt3 - 1}$$

But my calculator doesn't seem to understand what I'm saying. Is this ratio just totally wrong?

• Use $\tan^{-1}$. – Stephen Montgomery-Smith Nov 23 '13 at 1:53
• @StephenMontgomery-Smith facepalm It's 4 in the morning where I am. I guess I needed to write something stupid on the internet to realize I've been up too long! Thank you – Siyanda Nov 23 '13 at 1:56
• BTW, the quantity from which you want to calculate the $\;arccos\;$ is greater than $\;1\;$, so no wonder your calculator writes ERROR: that function's defined only for value in $\;[-1,1]\;$ ... – DonAntonio Nov 23 '13 at 1:56
• @StephenMontgomery-Smith : I'm wondering why you're not mentioning that one can subtract arguments. That seems like the main thing one needs to be aware of here. – Michael Hardy Nov 23 '13 at 2:01
• @MichaelHardy I didn't think of it. But that's a good point. I was merely trying to help the OP realize they had made an easy mistake. – Stephen Montgomery-Smith Nov 23 '13 at 2:04

You can subtract arguments. The argument of $-1+\sqrt{3}\,i$ is an angle in the second quadrant whose tangent is $\sqrt{3}/(-1)$, so it is $2\pi/3$ or $120^\circ$. The argument of $2+2i$ is an angle in the first quadrant whose tangent is $2/2$ so it is $\pi/4$ or $45^\circ$.
Subtracting, you get $120^\circ-45^\circ$ or $2\pi/2 - \pi/4$.
using eulers equation$$e^{i\theta} =\cos \theta+i\sin \theta$$the given expression can be written as $$\frac{\frac{-1}{2}+\frac{\sqrt{3}}{2}i}{1+i}$$which is$$\frac{e^{i\frac{2\pi}{3}}}{ e^{i\frac{\pi}{4}}}$$you get the resulting arguement$\frac{5\pi}{12}$