Angle between two complex vectors

Question

I'm trying to find the angle between $p = 2-2i$ and $q = 1-i\sqrt(3)$. I already got the answer as $\frac{\pi}{12}$ by subtracting the angles of $p$ and $q$ ($\frac{\pi}{3}-\frac{\pi}{4}$). I am trying to do the same using dot product, but Im not sure how to go about it.

Answer 1

Here is how to find the angle with the vector dot product. The two vectors are,

$$\vec{p}=(2,-2),\>\>\>\>\>\vec{q}=(1,-\sqrt 3)$$

Their dot product is

$$\vec{p}\cdot\vec{q} = pq\cos\theta$$

Then, $\cos\theta$ can be computed as below,

$$\cos\theta = \frac{\vec{p}\cdot\vec{q}}{pq} =\frac{2+2\sqrt 3}{\sqrt 8 \sqrt 4}=\frac{1+\sqrt 3}{2\sqrt 2}=\cos 15^\circ$$

Thus,

$$\theta =15^\circ$$

Answer 2

These are not complex vectors. They're still really since all the components are real. You may write each of them as $(2,-2),\,(1,-\sqrt 3).$ Then their dot product is $(2)(1)+(-2)(-\sqrt 3)=2+2\sqrt 3.$

Recall that for two vectors $a,b$ with an angular separation of $\phi,$ we have that $a\cdot b=|a||b|\cos\phi.$ Can you now continue?