Find angle $\alpha$ from a complex vector

Question

I'm trying to solve this problem from a Russian book:

Find the angle which is needed to rotate the vector $3\sqrt{2} + i2\sqrt{2}$ to obtain the vector $-5+i$.

EDIT: $\tan\dfrac{\pi}{6} \neq \dfrac{2}{3}$, so I'm wrong in most of what I've done.

Is there another way to solve this (I did this the wrong way :P )?

Any help will be appreciated. Thanks! :)

Answer 1

Hint: Divide $-5+i$ by $(3+2i)\sqrt2$.

Answer 2

The answer is $$ \arg(\frac{-5+i}{3+2i}) = \arg((-5+i)(3-2i)) = \arg(-13+13i) = \arg(-1+i)= 3\pi/4 $$