Find $z$ such that $|z+3-i√3|=√3$ and $z$ have the smallest argument

Question

I let $z$ equal to $(-3)$ so $|-i√3|=√3$ which is true but I don't know how to explain it or maybe this is not right at all. I mean maybe there is another solution that have the argument smaller than $z=-3$. Please help me

Answer 1

Let $z = x+iy$. Then, $$|z+3-i\sqrt{3}| = |(x+3)+i(y-\sqrt{3})| = \sqrt{(x+3)^2+(y-\sqrt{3})^2}$$

Now, setting this equal to $\sqrt{3}$ and squaring both sides: $$(x+3)^2+(y-\sqrt{3})^2=3$$

This is the equation of a circle, centre $(-3, \sqrt{3})$ and radius $\sqrt{3}$.

The question then becomes: Which point on this circle has the smallest argument?