Complex Numbers: How do I solve for relation between two complex numbers? Prove by contradiction that if $w,z\in\mathbb{C}$ such that $|w|\leq 1$ and $w^nz+w^{n-1}z^2+...+wz^n=1$, then $|z|>(1/2)$.
Any help on how to approach the problem will be helpful. I don't have any lead whatsoever! Thank you.
 A: The question asks you to prove a problem via contradiction.  So, I would first assume that $|z|\leq (1/2)$.  Then, take the absolute value of each side of your equation, giving: $$|w^nz+w^{n-1}z^2+...+wz^n|=|1|=1$$
Now, we have an absolute value on the left hand side, so we can apply the triangle inequality: $$|w^nz|+|w^{n-1}z^2|+...+|wz^n|\geq |w^nz+w^{n-1}z^2+...+wz^n|=1$$
Finally, we have that $|w|\leq 1$.  The above sum will clearly be maximized when $|w|=1$, so we can say: \begin{align}1\leq |w^nz|+|w^{n-1}z^2|+...+|wz^n|&\leq |1|^n|z|+|1|^{n-1}|z|^2+...+|1||z|^n \\&\leq |z|+|z|^2+...+|z|^n \end{align}
Finally, we can appy the formula for a finite sum of a geometric series.  First putting this series into "Standard Form" (meaning $1+r+r^2+...$), we get that $$1\leq |z|\left(1+|z|+|z|^2+...+|z|^{n-1}\right)$$
We have per the geometric sum formula that $$S_n=\sum_{k=0}^na_k=\sum_{k=0}^nr^k=\frac{1-r^{n+1}}{1-r}$$
Applying this formula with $r=|z|$ and $n=n-1$ (confusing notation here), we get that: $$1\leq |z|\frac{1-|z|^n}{1-|z|}\implies 1-|z|\leq |z|-|z|^{n+1}\implies1\leq 2|z|-|z|^{n+1} $$
Now we get to the point where we can apply $|z|\leq (1/2)$.  Consider the size of each of the terms on the right hand side if this is true: $$1\leq \underbrace{2|z|}_{\leq 1}-\underbrace{|z|^{n+1}}_{\leq (1/2)}$$
Clearly both of these numbers are strictly positive.  So, we get that $$1\leq 2|z|-|z|^{n+1}\leq 1-\epsilon$$
For some small, strictly positive $\epsilon$.  This is a contradiction.
A: By the triangle inequality we have:
$1=|w^nz+w^{n-1}z^2+\dots + wz^n|\leq |w^nz|+|w^nz^2|+\dots + |wz^n|=|w|^n|z|^1+|w|^{n-1}|z|^2+\dots + |w||z|^n$
Since $|w|\leq 1$ this is smaller than or equal to:
$|z|^1+|z|^{2}+\dots + |z|^n$. If $|z|\leq 1/2$ then this is smaller than or equal to:
$(1/2)+(1/2)^2+\dots + (1/2)^n = \frac{1-(1/2)^{n+1
}}{1/2}-1<2-1=1$.
Joining all inequalities we have $1=|w^nz+w^{n-1}z^2+\dots + wz^n|<1$, the contradiction comes from assuming $|z|\leq 1/2$. We conclude $|z|>1/2$.
A: The problem states "proof by contradiction" so you automatically have a lead: start by assuming the opposite of what you want to prove. In this case $|z| \leq \frac{1}{2}$. The contradiction will come from applying the triangle inequality to $|w^nz+w^{n-1}z^2+...+wz^n|=|1|=1$.
