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Let $f : \mathbb{C} \to \mathbb{C}$ $$f(x)=x^2 + 2x + 1$$ Prove whether f is a bijection or not.

This is what I have so far

Let $x_1,x_2 \in \mathbb{C}$. suppose $f(x_1)=f(x_2)$. $$x_1^2 + 2x_1 +1 = x_2^2 + 2x_2 + 1$$

user26857
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momsSpeghetti
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    What do you want to prove and what have you tried so far? And, of course, welcome on math.stackexchange :) – Severin Schraven Apr 03 '22 at 03:30
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    It might help, if you notice that $f(x)=(x+1)^2$. – Severin Schraven Apr 03 '22 at 03:33
  • i'm trying to prove that $x^2 + 2x + 1$ is a bijection. meaning it is onto and one-to-one. – momsSpeghetti Apr 03 '22 at 03:33
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    It is hard to prove, as it is not true :) it is onto, but it is not one-to-one. Try to find $x,y\in \mathbb{R}$ such that $(x+1)^2 = (y+1)^2$ and $x\neq y$. – Severin Schraven Apr 03 '22 at 03:34
  • I assume that it wasn't true. the question actually reads In each of the following parts, determine whether the given function is one-to-one and whether it is onto. – momsSpeghetti Apr 03 '22 at 03:36
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    Then you should indicate this in the question. We do not know what your exercise precisely is. Also you wrote above that you are trying to show that it is a bijection. In general it helps to first look at the graph of the function over the reals. Then you get a first impression of what the function might behave like. – Severin Schraven Apr 03 '22 at 03:38

2 Answers2

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$f: \Bbb C\to \Bbb C$ given by $f(x) = x^2 + 2x + 1$ is not one-one. To see this, it is enough to find $a,b\in \Bbb C$ such that $a\ne b$ and $f(a) = f(b)$. There are infinitely many ways to choose $a,b \in \Bbb C$ in the desired manner, and one of them is $$a = -2 \quad \text{ and }\quad b = 0$$ As a result, it is clear that even the restriction map $f\vert_\Bbb R: \Bbb R\to \Bbb R$ is not one-one.

$f$ is not a bijection.

Edit: This edit is in response to the comment below. Indeed, $f$ is onto. This is not hard to see. Let $z\in \Bbb C$. We want to find $y \in \Bbb C$ such that $f(y) = z$, i.e., $(y+1)^2 = z$. The choice $y = z^{1/2} - 1$ works. Note that $z^{1/2}$ can possibly take two different values in $\Bbb C$, and any of the two will work.

stoic-santiago
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For what it's worth, let $f$ be any complex polynomial. Then

  1. $z\mapsto f(z)$ is injective on $\mathbb C$ if and only if $\deg(f)=1$;
  2. $z\mapsto f(z)$ is surjective on $\mathbb C$ if and only if $f$ is not a constant polynomial.

(It follows that $z\mapsto f(z)$ is bijective if and only if $f$ is a linear polynomial.)

For the injectiveness problem, clearly $f$ is injective when $\deg(f)=1$ and not injective when $f$ is a constant polynomial. It remains to consider the case where $n=\deg(f)\ge2$. If $f$ has at least two distinct zeroes $z_1$ and $z_2$, then $f(z_1)=0=f(z_2)$ and $f$ is not injective. If $f(z)=0$ has only one root $a$ of multiplicity $n=\deg(f)$, then $f(z)=c(z-a)^n$ for some $c\ne0$. Therefore $f(a+e^{2\pi i/n})=c=f(a+1)$ and $f$ is not injective.

For the surjectiveness problem, if $f$ is a constant polynomial, then the image of $f$ is a single point and hence $z\mapsto f(z)$ is not surjective. If $f$ is not a constant polynomial, then for every $c\in\mathbb C$, $f(z)=c$ is solvable (with $n$ roots, in fact) by the fundamental theorem of algebra. Therefore $z\mapsto f(z)$ is surjective.

Ramen Nii-chan
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