A triangle and its median in complex plane. Let $z_1$, $z_2$, $z_3$ be vertices of the triangle $\triangle ABC$. And given that $|z_1|=|z_2|=|z_3|$. Then the median through $A$ cuts the circumcircle at which point? 
We need to get the answer in terms of the complex numbers $z_1$, $z_2$, $z_3$.
Any hints would be helpful.
 A: Hints:
Note that, since $z_1$,$z_2$, and $z_3$ all have equal magnitude, then they all lie on a common circle centred at the origin. Observe that this circle is the circumcircle of your triangle.
Now just create a line which goes through the point A and the mid-point of the opposite side.
Can you continue from here?
A: Let's simplify and particularize this problem by dividing the three complex numbers by $-z_1$, giving us $-1$ (let's call it $w_1$), $w_2=-\dfrac{z_2}{z_1}$, and $w_3=-\dfrac{z_3}{z_1}$. These three points give us a similar triangle to the original, rotated around the origin and shrunk so that the three new points are on the unit circle. When we find the $n$, the intersection of the median of this triangle and the unit circle, we can multiply that by $-z_1$ to get the answer to the original problem. So let's find $n$ in terms of $w_2$ and $w_3$.

The relevant median goes through the point $-1$ and the midpoint of $w_2$ and $w_3$ which is
$$m=\frac{w_2+w_3}2=-\frac{z_2+z_3}{2z_1}$$
By high-school geometry we see that the central angle for $n$ is twice the inscribed angle for $m$ relative to the point $-1$. $n$ is the point with that double-sized angle (argument for complex numbers) and distance $1$ from the origin (modulus $1$ for complex numbers). The final answer to the problem is $-z_1\cdot n$, or

$$-z_1\operatorname{cis}\left(2\arg\left[1-\frac{z_2+z_3}{2z_1}\right]\right)$$

or, if the function $\operatorname{cis}(\phi)=\cos\phi+i\sin\phi$ is not familiar,

$$-z_1e^{2i\arg\left(1-\frac{z_2+z_3}{2z_1}\right)}$$

These are the "simplest" forms of the answer in the sense that only one variable is repeated in the formula. We can use a more familiar complex function like conjugate or modulus if we use the facts that
$$\operatorname{cis}(\arg[u])=\frac{u}{|u|}$$
and 
$$\operatorname{cis}(2\arg[u])=\frac{u}{\overline u}$$
and get these other expressions for the answer:

$$-\overline z_1\cdot\frac{(z_2+z_3-2z_1)^2}{|z_2+z_3-2z_1|^2}$$

and

$$-\overline z_1\cdot\frac{z_2+z_3-2z_1}{\overline z_2+\overline z_3-2\overline z_1}$$

