Suppose $z_1, z_2,... z_n$ are in a disc $D$. Prove that the polynomial $f(z)=z^n-z_1z_2...z_n$ has a root in $D$. I have considered these cases: Let $C= \partial D$. If any $z_i = 0$, then let $z = 0$. If not, suppose the origin is in D, then $f'(z)/f(z)$ has zeroes in $D$ and $\int_C f'(z)/f(z) = 2\pi i $ (# zeroes - poles). By Argument principle, if $f(z)$ has no zeroes in $D$ then the integrand is holomorphic, which gives a contradiction. I am stuck at the case where $D$ does not contain the origin. If $f(z)$ has no zeroes in $D$, then $f′(z)/f(z)$ is nonzero on $D$, but I am not sure how to translate that to $\int_C f'(z)/f(z) \neq 0$.
 A: The problem appears to be from a “Romanian TST 2004” contest. The following proof by induction is essentially taken from Geometric mean for n complex numbers. We can assume that all $z_j$ are non-zero because the statement is trivially true otherwise.
I will also assume that $D$ is an open disk, but the same proof works with small modifications also for a closed disk.
The case $n=1$ should be clear, so let us assume that $n \ge 2$ and the statement has been proven for all smaller values of $n$.
Assume that $f(z) = z^n - z_1z_2 \cdots z_n$ has no zero in $D$, i.e.
$$
 f(z) = (z-u_1) (z-u_2) \cdots (z-u_n)
$$
with $u_1, u_2, \ldots , u_n \notin D$. Define
$$
 g(z) = (z_1-z)f'(z) + nf(z) = n z_1 (z^{n-1} - z_2 z_2 \cdots z_n) \,.
$$
From the induction hypothesis we know that $g(u) = 0$ for some $z\in D$. Then
$$
 \frac{1}{u-z_1} = \frac 1n \frac{f'(u)}{f(u)} = \frac 1n \left( \frac{1}{u-u_1} + \frac{1}{u-u_2} + \cdots + \frac{1}{u-u_n}\right) \, .
$$
The function $\phi(w) = 1/(u-w)$ maps the disk $D$ to the exterior of some disk $E$, and the complement of $D$ to $\overline E$. Then
$$
 \phi(z_1) = \frac 1n \bigl( \phi(u_1) + \phi(u_2) + \cdots + \phi(u_n)\bigr)
$$
is a contradiction, because the left-hand side is a point in the exterior of $E$, and the right-hand side is a convex-combination of points in $\overline E$.
A: I will solve it only in the case $D$ does not contain the origin. By a complex homotethy, we may reduce to the case $D$ is the unit disk translated by distance $d> 0$. Now, there exists a determination of the $\log$ on $D$. Let us show that the image under $\log$ of $D$ is convex. For this, it is enough to show that the curve $\log\circ \gamma$ has everywhere positive curvature, where $\gamma$ is the boundary of $D$. Now, we  use the formula for curvature of the image of the unit disk under the holomorphic map $f(z)$.
$$\rho_{f(z)} = \frac{1}{|zf'(z)|} \text{Re}\bigg(1 + \frac{zf''(z)}{f'(z)}\bigg).$$
To show that the curvature is positive, it is enough to show that the expression in the bracket has positive real part. But for $f(z) = \log(z+d)$ we get
$$1 + \frac{zf''(z)}{f'(z)} = \frac{d}{z+d}$$
has positive real part for $|z|=1$ and $d>1$.
Now all is ready. Let $z_1$, $\ldots$, $z_n$ in $D$. Consider $D' = \log D$, where $\log$ is a determination of the logarithm on $D$. Consider $z'_k = \log z_k$. Since $D'$ is convex, we have
$$z'\colon =\frac{\sum_{k=1}^n z'_k}{n} \in D'$$
Now take $z= \exp z' \in D$. We have $z^n = z_1  \cdots z_n$.
