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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.

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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 $z=z'$.

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  • $\begingroup$ 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. $\endgroup$ – Sun Joong Kim Nov 26 '19 at 14:52
  • $\begingroup$ 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. $\endgroup$ – Bernard Nov 26 '19 at 18:37
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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 $\operatorname{Re}(z) < 2$.

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

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