Question:
For:$$|z - z_1|^2+|z - z_2|^2+|z - z_3|^2+\cdots+|z - z_n|^2 = S$$ Prove that the minimum value of $S$ is when:$$z = \frac{z_1+z_2+z_3+\cdots+z_n}{n}$$
I have no idea how to even start this question. I tried to do it graphically but that didn't get me anywhere. However, I feel that the easiest proof for this would be graphically.
In fact we have following identity,For any complex numbers $z_{1},z_{2},\cdots,z_{n}$,we have $$\dfrac{1}{n}\sum_{j=1}^{n}|z-z_{j}|^2=n\left|z-\dfrac{z_{1}+z_{2}+\cdots+z_{n}}{n}\right|^2+\sum_{j=1}^{n}\left|z_{j}-\dfrac{z_{1}+z_{2}+\cdots+z_{n}}{n}\right|^2$$ so when $$z=\dfrac{z_{1}+z_{2}+\cdots+z_{n}}{n}$$ then $$\sum_{j=1}^{n}|z-z_{j}|^2$$ is minimum.and the value is $$n\cdot\sum_{j=1}^{n}\left|z_{j}-\dfrac{z_{1}+z_{2}+\cdots+z_{n}}{n}\right|^2$$
