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Let $a$ and $b$ be some complex numbers and let be $f: \mathbb{C} \to \mathbb{C}$ such that $f(z)=az + b\overline{z}$. Prove that $f$ is bijective if and only if $|a|-|b|\neq 0$.

My idea was to find another function $g: \mathbb{C} \to \mathbb{C}$ with the property $g \circ f = f \circ g= \mathbb{1}_{\mathbb{C}}$ and to use the condition $|a| \neq |b| \Leftrightarrow |a|^2 \neq |b|^2$ but I do not know how to continue. Could somebody help me?

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If $|a|=|b|$ then $az+b\overline {z}=a0+b\overline {0}$ where $z=-b\frac {\overline {z}} {a}$ (if $a \neq 0$ and any $z$ if $a=0$) so $f$ is not one-to one. [Note that $z=-b\frac {\overline {z}} {a}$ holds if $z$ is one of the square roots of $-\frac b a$].

If $|a|\neq |b|$ then $az+b\overline {z}=aw+b\overline {w}$ implies $a(z-w)=-b(w-z)$. Hence $|a(z-w)|=|-b(w-z)|$ which implies $w=z$.

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  • $\begingroup$ but if i choose z the square root of -b/a and 0 , it is clearly that f (square root of -b/a) = f(0) and f is not bijective why is it necessary that |a|=|b| to make this artifice $\endgroup$ – mathematiciangrade8 Feb 6 at 8:32
  • $\begingroup$ @mathematiciangrade8 The fact that $|z|=1$ (which implies $\overline {z}=\frac 1 z$) when $z$ is one of the square roots is required in this argument. That is why we need $|a|=|b|$. $\endgroup$ – Kavi Rama Murthy Feb 6 at 8:36
  • $\begingroup$ thank you sir ! $\endgroup$ – mathematiciangrade8 Feb 6 at 8:40
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Your function is $\mathbb{C}$-linear. You can look at $f$ as an $\mathbb{R}$-linear function $f:\mathbb{R}^2\to \mathbb{R}^2$ and then bijectivity reduces to having determinant of the matrix different from zero. The determinant is precisely $|a|^2-|b|^2$.

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