Animated Solution for Mathematics - Complex Numbers: Prove that there exists no complex number $z$ such that $|z|<1/3$ and ${\sum }_{r=1}^n{a}_r{z}^r=1$ where $|a_r|<2$.

Imagine you are standing on the complex plane, looking at a tiny, restricted region. This is the domain of our problem: $|z|<\frac{1}{3}$.

Any complex number $z$ we consider must live strictly inside this fence. It cannot touch the boundary; it is trapped in this small circle.

Our goal is to see if a polynomial sum ${\sum }_{r=1}^n{a}_r{z}^r$ can ever reach the value of $1$, given that the coefficients satisfy $|a_r|<2$.

The Triangle Inequality

To tackle this, we need a weapon: the Triangle Inequality. It states that the magnitude of a sum is always less than or equal to the sum of the individual magnitudes:

Think of this as the shortest path principle. By taking the absolute value of the entire sum, we are essentially looking for the "worst-case scenario"—the maximum possible magnitude this sum could ever achieve.

Our Secret Weapon

We know that $|z^r|=|z{|}^r$. Substituting our given bounds $|a_r|<2$ and $|z|<\frac{1}{3}$ into the inequality, we obtain:

Factoring out the constant $2$, we are left with $2{\sum }_{r=1}^n(\frac{1}{3}{)}^r$. This is a classic Geometric Progression (G.P.) with the first term $a=\frac{1}{3}$ and common ratio $r=\frac{1}{3}$.

Using the G.P. sum formula $S_n=\frac{a(1-r^n)}{1-r}$, we calculate the upper bound:

Simplifying the denominator $1-\frac{1}{3}=\frac{2}{3}$, the expression becomes:

The fractions cancel out beautifully, as $\frac{1/3}{2/3}=\frac{1}{2}$, and $2\cdot \frac{1}{2}=1$. We are left with the result:

Since $n\ge 1$, the term $(\frac{1}{3}{)}^n$ is always positive. Therefore, $1-(\frac{1}{3}{)}^n$ is strictly less than $1$.

We have proven that the magnitude of the sum is always strictly less than $1$. It can never reach the target. The sum is trapped, and the proof is complete.