Finding vertices of a triangle using complex numbers

The question wants me to find the coordinates of the other two vertices B and C.

What I did was that I converted A into complex coordinates, so 4-i.

Then I want to find vertex B, and I know that equilaterial triangle, each angle is $60 ^{\circ}$, so I did $(4-i)cis(\pi/3)$

The steps are as follows:
$(4-i)cis(\pi/3)=(4-i)[cos(\pi/3)+isin(\pi/3)]=(4-i)(\frac12+\frac{\sqrt3}{2}i)=2+2\sqrt(3)i-\frac12 i+\frac{\sqrt(3)}{2}$

So the coordinates of B becomes $(\frac{4+\sqrt3}{2}, \frac{4\sqrt3-1}2)$, but the answer given to this question says it's $(\frac{\sqrt3-4}{2}, \frac{4\sqrt3+1}2)$. Where did I go wrong?

To find vertex C, I have also used a similar method, but multiplying 4-i with $cis(-\frac{\pi}{3}). However, I have also gotten a slightly wrong answer involving the signs.

Please advise. Sorry in advance for any mistakes in labelling of the title and tags (I'm not very good at those, but am trying to improve on it)

Answer

Assuming that the center of the triangle is the origin $z=0$, to get from $A$ to $B$ you should rotate by $120$ degrees - or $2\pi /3$ radians.

To see this, connect the origin $O$ to both points $A$ and $B$ and try to see what the angle is between $OA$ and $OB$.