$f(z) = z^k + kz$ is injective in the unit disc

Question

How would I go around proving that $f(z) = z^k + kz$ in injective in the open unit disc in $\mathbb{C}$, for each natural $k$?

I don't really know where to start. I observed that its derivative is nonzero, but that was all that I came up with.

Any help is appreciated. Thanks.

More generally, if $\mathrm{Re},f'>0$ on a convex domain, then $f$ is injective. And you have $f'(z)=k(z^{k-1}+1)$. — 40 votes, Jul 20 '13 at 21:53

score 5 · Accepted Answer · answered Jul 20 '13 at 21:25

5

If $z_1^k+kz_1=z_2^k+kz_2$ then

$$(z_1-z_2)\left( z_1^{k-1}+z_1^{k-2}z_2+...+z_2^{k-1}+k\right)=0$$

Hint if $z_1,z_2$ are in the open unit disk, then by the triangle inequality

$$\left| z_1^{k-1}+z_1^{k-2}z_2+...+z_2^{k-1} \right| < k \,.$$

answered Jul 20 '13 at 21:25

N. S.

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