Prove that $\max _{z \leq 1} |az^n+b|=|a|+|b|, a,b \in \mathbb C$ Prove that $\max _{|z| \leq 1} |az^n+b|=|a|+|b|, a,b \in \mathbb C$
We'll get the maximum of the function $|f(z)|=|az^n+b|$ at $|z|=1$.
So what I tried doing is setting $z=e^{it}$, $a=r_1e^{i \alpha}, b=r_1e^{i\beta}$, plugged it in $|az^n+b|$.
Eventually, I got $\max _{|z| = 1} \sqrt{r_1^2+2r_1r_2cos(\alpha+nt)+r_2^2}$. It would be very comfortable to say we get the max of that expression when $\alpha+nt=0$, and then get desired result - but I couldn't justify this - for instance when $r_1$ or $r_2$ are negative...
Any assistance will be great!    
 A: by triangle inequality, you have $$|az^n + b| \le |a||z|^n+|b| \le  |a| + |b| \text{ if } |z| \le 1.$$  therefore $$   \max _{z \leq 1} |az^n+b|=|a|+|b|, a,b \in \mathbb C$$
$\bf p.s.$ the equality occurs when $|z| = 1$ and $arg(az^n)= arg(b).$  if $a = 0,$ then there is nothing to do.
in the case $a \neq 0,$  we have $$\arg(a) + n \arg(z) = \arg(b) \to \arg(z) = \frac{\arg(b) - \arg(a)}{n} $$ so that $$z^n = e^{i(\arg(b) - \arg(a))}, az^n = |a|e^{i\arg(b)}, az^n + b = (|a|+|b|)e^{i\arg(b)}$$ and finally $$|az^n + b| = |a| + |b| \text{ where } z = e^{i\left(\frac{\arg(b) - \arg(a)}{n} \right) }.$$
A: Of course, we always have $\max_{|z| \le 1} |a z^n + b| \le |a| + |b|$ from the triangle inequality.
To see what exactly is the maximum, let's simplify a bit. If $a = 0$ or $b = 0$ the result is obvious, so assume $a \ne 0$ and $b \ne 0$. Denote $c = -\frac{b}{a} \ne 0$, then   
$$
\begin{eqnarray*}
\max_{|z| \le 1} |a z^n + b| &= |a| \cdot \max_{|z| \le 1} |z^n - c| \\
&= |a| \cdot \max_{|z'| \le 1} |z' - c|
\end{eqnarray*}
$$
The second equality is just a change of variable $z' = z^n$. Now we're looking for the point $z_0$ in the unit disc $D = \{z, \, |z| \le 1\}$ that is the farthest from $c$. If you draw a picture, you'll see that point is in the intersection of the unit circle $C = \{z, \, |z| = 1\}$ with the line joining $0$ and $c$ "diametraly opposite" to $c$ (there are two such points, but one of them is clearly further from $c$) and that $|z_0-c| = 1 + |c|$. So we get as expected
$$\max_{|z| \le 1} |a z^n + b| = |a| \cdot (1+|c|) = |a|+|b|$$
