show that $\max_{|z|\le 1}|az^n+b|=|a|+|b|$

Question

Can some one tell me why $\max_{|z|\le 1}|az^n+b|=|a|+|b|$,where z is unit circle of the complex plane.

I know that based on maximum principle theorem, the largest value should be in the boundary $|z|=1$, and why there exits a z in the boundary such that $|az^n+b|=|a|+|b|$?

Answer 1

The triangle equality is an equality when both variables are parallel (and, to avoid a common linguistic ambiguity, not antiparallel). Once $|z| = 1$, all $z^n a$ does is rotate $a$. Some of those specific rotations make $a$ and $b$ parallel and those give you the maximum of your expression.

If we go in the other direction, letting the angle vary first, the maximum will still occur when $z^n a$ and $b$ are parallel. Then letting $|z|$ increase to one, the sum increases (from $b$ plus a little parallel copy of $a$ to $b$ plus a full-sized parallel copy of $a$).

Answer 2

Assuming $|z|=1$, using the triangle inequality you'll get $$|az^n+b|\leq|a||z|^n+|b|\leq|a|+|b|$$ Now if you find some $z$ for which $|az^n+b|=|a|+|b|$, then you can say $$\max_{|z|\leq1}|az^n+b|=|a|+|b|$$ And that $z$ could be ($a\ne 0$) $$z=\left(\frac{|a|+|b|-b}{a}\right)^\frac 1n$$