Geometry Complex Numbers Convexity

Question

I'm struggling with the following problem,

Let $g(z)=\sum^k_1 m_{\alpha}(z-z_{\alpha})^{-1}$. Show that if $g(z)=0$, then $z_1,\cdots,z_k$ cannot all lie on the same sie of a straight line through $z$.

What I did:

The book says that I should use the fact that if $z_1,\cdots,z_k$ lie on one side of some straight line through 0, then $z_1+\cdots+z_k\neq 0$. But I can't move much, I think it follows that the $m_{\alpha}(z-z_{\alpha})^{-1}$'s are on the same side of a line through zero, I know that the next step is probably going to be something related to expanding this thought to the inverse of these numbers and the last one is likely going to be translating $z-z_\alpha$ to $z$.

Answer 1

If $z_1,\ldots,z_n$ lie on the same side of a line through $z$, then $z-z_1,\ldots,z-z_n$ lie on the same side of a line through $0$. Let this line pass through $0$ and the complex number $w$. Then $(z-z_1)/w,\ldots,(z-z_n)/w$ lie on the same side of the real axis. Then $(z-z_j)/w=a_j+b_ji$ where the $b_j$ are all positive or all negative.

What does this tell you about $\sum_j m_j/(z-z_j)$?

Answer 2

Assuming $m_k > 0$

$$ (z-z_k)^{-1} = \frac{x-x_k-i(y-y_k)}{\vert z-z_k\vert^2} $$

then

$$ \sum_k\frac{m_k(x-x_k)}{\vert z-z_k\vert^2} = \sum_k\frac{m_k(y-y_k)}{\vert z-z_k\vert^2} = 0 $$

then

$$ x^* = \frac{\sum\frac{m_kx_k}{\vert z-z_k\vert^2}}{\sum\frac{m_k}{\vert z-z_k\vert^2}} $$

and analogously for $y^*$

Now

$$ \sum_k\frac{m_k(z_k-z^*)}{\vert z^*-z_k\vert^2} = 0\Rightarrow z_k \ \ \mbox{cannot be located only into one of the semi-planes defined by } z = \lambda z^* $$