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We consider $T:\mathbb{C}\to\mathbb{C}$ defined by $T(z)=\lambda z+\mu\overline{z}$, where $\lambda ,\mu\in\mathbb{C}$. I want to prove that if $T$ is injective, then $\lambda\cdot\overline{\lambda}\neq\mu\cdot\overline{\mu}$.

Clearly, if $\mu =0$, then $\lambda\neq 0$ because $T$ is injective.

Now, if $\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}$ and $\mu\neq 0$, then I wanted to find a $z\in\mathbb{C}\setminus\{0\}$ such that $z=-\dfrac{\overline{\lambda}}{\overline{\mu}}\overline{z}$, because we would have that:

$T(z)=-\lambda\dfrac{\overline{\lambda}}{\overline{\mu}}\overline{z}+\mu\overline{z}=\overline{z}(-\lambda\dfrac{\overline{\lambda}}{\overline{\mu}}+\mu)=0$

('Cause this would prove that $\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}$ implies $T$ is not injective.)

So, the question is, can we always find such a $z\in\mathbb{C}\setminus\{0\}$?

Michael Albanese
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Talexius
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    This follows directly from the result at http://math.stackexchange.com/q/471634/ – Jonas Meyer Aug 25 '13 at 20:35
  • T. Bongers, you're right. So it seems my "solution" is not correct. – Talexius Aug 25 '13 at 20:44
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    Just a sec! We suppose that $\lambda\cdot\overline{\lambda}=\mu\cdot\overline{\mu}$, so $|\lambda|=|\mu|$... – Talexius Aug 25 '13 at 20:52
  • Could you edit your question accordingly? – Michael Albanese Aug 25 '13 at 21:05
  • @MichaelAlbanese: Does the question need to be edited? Talexius's comment about the work being incorrect was incorrect (which seemed to be noted in the subsequent comment). I don't know what T. Bongers saw as problematic. – Jonas Meyer Aug 25 '13 at 21:09
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    @JonasMeyer: I see, I thought that Talexius had written the question incorrectly and was commenting on what he was supposed to ask. Instead, he was replying to T.Bongers. – Michael Albanese Aug 25 '13 at 21:12
  • Talexius: For future reference, if you write @username to reply to username, then that user gets alerted of your response. Otherwise they will only see it if they happen to check back and read the comments. (You get alerted of comments on your own posts regardless, but for other commenters the @ symbol is needed.) Using this format also makes it clearer to others that your comment is a response to something in particular. – Jonas Meyer Aug 25 '13 at 21:16
  • @JonasMeyer, I see, thanks. – Talexius Aug 25 '13 at 21:31

3 Answers3

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View $T$ as an $\Bbb R$-linear map.

Since $T(1)=\lambda+\mu$ and $T(i)=\lambda-\mu$ the determinant of $T$ is $$ \det(T)=(\lambda_1+\mu_1)(\lambda_1-\mu_2)-(\lambda_2+\mu_2)(-\lambda_2+\mu_2)= \lambda_1^2+\lambda_2^2-\mu_1^2-\mu_2^2=|\lambda|-|\mu|, $$ where the subscript 1 denotes real part and the subscript 2 denotes imaginary part. Therefore $T$ is injective if and only if $$ |\lambda|\neq|\mu|. $$

Note : the first version of this answer contained an error of computation. This justifies the comment of Jonas Meyer.

Andrea Mori
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  • Once the computation is fixed this gives a nice proof that injectiveness is equivalent to $|\lambda|\neq|\mu|$. – Jonas Meyer Aug 25 '13 at 21:28
  • I think you wanted to write $|\lambda|\neq |\mu|$ – Talexius Aug 25 '13 at 21:30
  • @Talexius: Thanks, fixed. – Jonas Meyer Aug 25 '13 at 21:31
  • @AndreaMori: I chose Meyer's answer because he answered first, but I like your answer as well. Thanks. – Talexius Aug 25 '13 at 21:41
  • @Talexius: No problem. Just because you seem new to the site, a good advice is to wait several hours to acknowledge an answer so to give others a motivation to come up with alternative solutions. Nobody's going to complain if you do not acknowledge immediately. – Andrea Mori Aug 25 '13 at 21:46
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Here's an alternative proof.

Suppose that $T$ is injective. Note that $\mu \neq \bar{\lambda}$, otherwise $T(z) = \lambda z + \overline{\lambda z}$ in which case $T(z) = 0$ for any $z \in \mathbb{C}$ such that $\Re(\lambda z) = 0$. Now note that $T(\bar{\lambda}) = |\lambda|^2 + \mu\lambda$ and $T(\mu) = \lambda\mu + |\mu|^2$. Therefore $T(\bar{\lambda}) - T(\mu) = |\lambda|^2 - |\mu|^2$; as $\mu \neq \bar{\lambda}$ and $T$ is injective, $|\lambda|^2 \neq |\mu|^2$.

Michael Albanese
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To "generalize" your reformulation to an equivalent statement: Given $a\in\mathbb C$ with $|a|=1$, can we find $z\in\mathbb C\setminus\{0\}$ such that $z=a\overline z$?

The answer is yes. E.g., restricting to $|z|=1$, we can reformulate this as finding a solution $x\in\mathbb R$ to the equation $e^{ix}=e^{i\theta}e^{-ix}$, which is easy.

Jonas Meyer
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