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I am reviewing conformal mappings for a complex analysis exam, and I can't get my head around this question. I have some $a\in D(0;1)$, and for $z\in D(0;1)$ let $$f(z)=\frac{a-z}{1-\overline{a} z}$$ I have successfully shown, using the fact that the derivative of a conformal map must be non-zero everywhere, that this map f is conformal. But how can I show, that this is a bijection from $D(0;1)$ to itself? Do I show that it maps the boundary of $D(0;1)$ onto the boundary of $D(0;1)$ and if so, how do I do this?

Any help really appreciated.

Thanks.

AJY
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xyz12345
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3 Answers3

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If it's conformal, it's bijective, so it suffices to show that it sends $D\rightarrow D.$ If $z\in D$, then $$|f(z)|^2=\frac{|z|^2+|a|^2-z\bar{a}-\bar{z} a}{1+|z|^2|a|^2-z\bar{a}-\bar{z}a}.$$ Recall that $|z|^2<1.$ Multiplying this by $1-|a|^2,$ we get that $|z|^2-|a|^2|z|^2<1-|a|^2,$ which implies that $|f(z)|^2<1.$

EDIT: The phrasing in the question is somewhat confusing, with the statement saying that it's conformal, which seems to be assuming what we want to prove in some sense. I've shown that it maps $D$ to $D$. As @Maxim pointed out, I should say something about bijectivity. In this case, it is clear because $f$ is an involution from $D$ to $D$, and hence it's bijective.

cmk
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  • $f$ is bijective locally, but by itself this doesn't prove that $f: D \to D$ is surjective and doesn't prove that $f: D \to D$ is injective. – Maxim Jun 04 '19 at 14:39
  • If you take conformal to mean analytic and bijective between two sets, why do you need to estimate $|f(z)|^2$? $f: D \to D$ is conformal automatically means $f: D \to D$ is bijective. If you take conformal to mean analytic and bijective only locally around each point (this is what is implied by a non-zero derivative), then you need to prove that you get all of $D$, not a proper subset, and you need to prove that you don't get some points more than once (consider $z^2$ on an annulus around the origin). – Maxim Jun 04 '19 at 15:13
  • I'm not saying what you have is incorrect, just that it doesn't answer the question. You take for granted that a Mobius transformation is injective, but even then, $|f(z)|^2 < 1$ does not show that $f$ maps $D(0, 1)$ to $D(0, 1)$ and not, say, to $D(0, 1/2)$. – Maxim Jun 05 '19 at 15:17
  • My issue is with the first part of your last comment. Suppose I say "$z \mapsto z/2$ sends any point in $D$ to a point in $D$ and is bijective, therefore it sends $D$ to $D$". The antecedent is true if "bijective" means "bijective between $D$ and the image of $D$". Your answer can be repeated verbatim for $z \mapsto z/2$. Therefore something is missing. – Maxim Jun 05 '19 at 16:51
  • @Maxim I’m going to delete my comments later on. Thanks for pointing out my mistake! – cmk Jun 05 '19 at 17:32
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$f$ is a Möbius transformation.

Let's take three points on the boundary of $D(0,1)$. Namely, $1,-1$ and $i$.

$f(1)=\dfrac {a-1}{1-\bar a}$. By symmetry (since $1$ is on the $x$-axis), $\mid f(1)\mid=1$.

Similarly, $f(-1)=\dfrac {a+1}{1+\bar a}$, so $\mid f(-1)\mid=1$, (since $a$ and $\bar a$ are symmetric wrt $-1$).

Next $f(i)=\dfrac {a-i}{1-i\bar a}$. Write $a=x+iy$. Then $f(i)=\dfrac{x+(y-1)i}{1-y-xi}$. Thus $$|f(i)|=\sqrt{\frac{x^2+(y-1)^2}{(1-y)^2+x^2}}=1$$

Thus $f$ maps $S^1$ onto $S^1$.

Finally note that $f(0)=a\in D$. So $f$ maps $D$ onto $D$.

Jacob Manaker
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Lets show that for $a \in D(0,1)$ the function $f(z)=\frac{a-z}{1-\overline{a}z}$ is bijective straight from the definition.

$f$ is injective: Suppose $f(z)=f(w)$

then $\frac{a-z}{1-\overline{a}z} =\frac{a-w}{1-\overline{a}w}$

So $(a-z)(1-\overline{a}w)= (a-w)(1-\overline{a}z)$

or $a - z - |a|^2w +\overline{a}zw=a-w-|a|^2z+\overline{a}zw$

which simplifies to $(z-w)+|a|^2(w-z)=0$

or equivalently, $(z-w)(1-|a|^2)=0 \implies z=w$

$f$ is surjective: Suppose that $w \in D(0,1)$ then it is easy to show (see cmk's answer) that if $f(z) = w$ has unique solution in $D(0,1)$.

Mustafa Said
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