# Is the function $f(z)=z^2-4z$ (on the unit disk) injective?

I was exploring examples of elementary complex functions, keeping in mind the fact that ($$*$$) injective holomorphic function has no zero derivative, provided that the function is defined on an open and connected subset of $$\mathbb C$$.

Injectivity of complex functions seems quite different from that of real functions. In the real case, $$f(x)=x$$ was injective, $$f(x)=x^2$$ was not, $$f(x)=x^3$$ was injective, all of which were obvious from the graphs of the functions or anything. The injectivity of $$f(x)=x^n$$ depended only on whether $$n$$ is odd or even.

In the complex case, where the domain is the open unit disk $$U$$, $$f(z)=z$$ is injective, $$f(z)=z^2$$ is not, $$f(z)=z^3$$ is not ($${z_1}^3={z_2}^3$$ if $$z_1=\frac12$$, $$z_2=\frac12e^{\frac23\pi i}$$). It seems like $$f(z)=z^n$$ fails to be injective for $$n\ge2$$. And these were actually obvious from ($$*$$), since $$0\in U$$ and $$f'(z)=nz^{n-1}$$.

So, in order to create some nontrivial injective holomorphic function defined on $$U$$, $$f'$$ must not have zero in U. Because I was exploring the most elementary cases(i.e. polynomials), and polynomials always have zero somewhere, I came up with the following complex polynomial $$f(z)=z^2-4z$$ where $$f'(z)=2(z-2)$$ and the zero of $$f'$$ is away from $$U$$.

But is it injective? Here is my attempt; $$\begin{gather*} f(z_1)=f(z_2)\\ {z_1}^2-4z_1={z_2}^2-4z_2\\ (z_1-z_2)(z_1+z_2-4)=0\\ z_1=z_2 \end{gather*}$$ From third line to fourth line, $$z_1+z_2\neq4$$ since they belong to $$U$$. I think my attempt was right, but please tell me whether it is right or wrong.

You can see it is injective on the unit disk directly:

Suppose $$z^3-12z=z'^3-12z'$$. This means $$z^3-z'^3-12z+12z'=(z-z')(z^2+zz'+z'^2-12)=0.$$ Now, since we're n the unit disk, the triangle inequality ensures that $$\;|z^2+zz'+z'^2|\le |z|^2+|z||z'|+|z'|^2 \le 3$$, so the second factor can't be $$0$$, and the only solution is $$z=z'$$.

• I'm sorry, the title was wrong. It should have been $f(z)=z^2−4z$. And in the similar fashion as you did, I made some answer above. I think it's right. Thanks a lot. – Sun Joong Kim Nov 26 '19 at 14:52
• Yes it is right. You just add the details why $z_1+z_2$ can't be $4$ when $z_1,z_2$ line in the unit circle. – Bernard Nov 26 '19 at 18:37

Your argument is correct. It shows that $$f(z) = z^2 -4z$$ is in fact injective in the larger disk with radius $$2$$ centered at the origin.

An alternative approach is to use that If $\operatorname{Re}f^\prime > 0$ on a convex domain, then $f$ is one-to-one.. Applied to $$g(z) = -f(z) = 4z - z^2$$ we have $$\operatorname{Re} g'(z) = \operatorname{Re} (4 - 2z) = 4 - 2 \operatorname{Re}(z)$$ which shows that $$g$$ (and consequently, $$f$$) is injective in the halfplane $$\operatorname{Re}(z) < 2$$.

• Thanks for your kind answer. So I could only check the sign of $\text{Re}(f')$ – Sun Joong Kim Nov 26 '19 at 12:23
• @SunJoongKim: It is one criterion for injectivity, which can be useful. In your case it is not better or worse than what you did. – Martin R Nov 26 '19 at 12:27
• I see, Moreover, I think the proof(your link) is quite easy to understand or memorize. Thanks a lot! – Sun Joong Kim Nov 26 '19 at 12:34