# Complex function property

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?

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 $$w=z$$.
• @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 $|a|=|b|$. – Kavi Rama Murthy Feb 6 at 8:36
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 $$|a|^2-|b|^2$$.