Problem source

Define the modulus of a complex number $z$ and give the geometric interpretation of $\vert\,z_1-z_2\,\vert$ for two complex numbers $z_1$ and $z_2$. On the basis of this interpretation establish the inequality
$$\vert\,z_1+z_2\,\vert\le \vert\,z_1\,\vert+\vert\,z_2\,\vert.$$
Use this result to prove, by induction, the corresponding inequality for $\vert\,z_1+\cdots+z_n\,\vert$.
The complex numbers $a_1,\,a_2,\,\ldots,\,a_n$ satisfy $|a_i|\le 3$ ($i=1, 2,  \ldots ,  n$). Prove that the equation $$a_1z+a_2z^2\cdots +a_nz^n=1$$
has no solution $z$ with $\vert\,z\,\vert\le 1/4$.

Solution source

Suppose $z = a+ib$, where $a,b \in \mathbb{R}$ then the modulus of $z$, $|z| = \sqrt{a^2+b^2}$. Noting the similarity to the Pythagorean theorem, we can say that $|z_1 - z_2|$ is the distance between $z_1$ and $z_2$ in the Argand diagram.

\begin{align*}
|z_1 + z_2| &= |(z_1 - 0) + (0 -z_2)| \\
&\underbrace{\leq}_{\text{the direct distance is shorter than going via }0} |z_1 - 0| + |0 - z_2| \\
&= |z_1| + |-z_2| \\
&= |z_1| + |z_2|

\end{align*}

Claim: $\displaystyle \vert\,z_1+\cdots+z_n\,\vert \leq \sum_{i=1}^n |z_i|$
Proof: (By Induction)

Base Case: $n = 1, 2$ have been proven.

Inductive step, suppose it is true for $n = k$, then consider $n = k+1$, ie

\begin{align*}
\vert\,z_1+\cdots+z_k+z_{k+1}\,\vert &\leq \vert\,z_1+\cdots+z_k\vert + \vert z_{k+1}\,\vert \\
&\underbrace{\leq}_{\text{inductive hypothesis}}  \sum_{i=1}^k |z_i| + |z_{k+1}| \\
&= \sum_{i=1}^{k+1} |z_i|
\end{align*}

Therefore if our hypothesis is true for $n = k$ it is true for $n = k+1$, and so since it is true for $n = 1$ it is true by the principle of mathematical induction for all integers $n \geq 1$.

Suppose $|z| \leq 1/4$, then consider:

\begin{align*}
 \vert a_1z+a_2z^2+\cdots +a_nz^n \vert &\leq \vert a_1 z\vert + \vert a_2z^2\vert + \cdots + \vert a_n z_n\ \vert  \\
&= \vert a_1\vert\vert z\vert + \vert a_2\vert\vert z^2\vert + \cdots + \vert a_n\vert\vert z^n\ \vert \\
&\leq 3\left ( |z| + |z|^2 + \cdots + |z|^n \right) \\
&\leq 3 \left ( \frac{1}{4} + \frac1{4^2} + \cdots + \frac{1}{4^n} \right) \\
&< 3 \frac{1/4}{1-1/4} \\
&= 1
\end{align*}

Therefore we cannot have equality and there are no solutions.