Mapping of a Disk onto a Half-Plane

Question

look at this text:

In example 3 The following map is discussed: $$w=-{{z+1}\over{z-1}}$$ The author claims that this linear fractional transformation maps the unit disk onto the right halfplane but does not prove it. How it can be proved?

here is the graph of this mapping that has been drawn with maple software:

[This text is on page 748 of ''Advanced Engineering Mathematics tenth edition'' by ''Erwin Kreyszig'']

Answer 1

The author claims that this linear fractional transformation maps the unit disk onto the right halfplane but does not prove it.

Well, it's obvious – almost: $$f:z\mapsto-\frac{z+1}{z-1} \tag 1$$ is a Möbius transformation which means it maps circles to circles on the Riemann sphere $\Bbb C\cup\{\infty\}$. In the complex plain there are special cases though, namely lines have to be ragarded as circles that run through $\infty$.

• The image of the unit circle $S$ must be such a line, because $1\in S$ and $f(1)=\infty$.

• To determine which line, observe that $f(-1)=0$ so the image is a line through the origin. And for $i\in S$ we have $f(i)=-(i+1)/(i-1)=-(i+1)^2/2=-i$. Thus, $f(S)$ is the imaginary line $g:\Re z = 0$.

• To find whether the interior is mapped to the left or to the right of $g$, we take the image of some point in the interior of $S$, say $z=0$ which $f$ maps to 1. Thus, the interior of $S$ maps tho the right half plane.

All this is a bit hand-waving, but perhaps more instructive than some fancy calculation.

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As a final treat, here is a nice animation about Möbius transforms.

Answer 2

Yes, this is a standard and important example.

To be completely rigorous, we should first prove that linear fractional (Moebius) transformations map lines-and-circles to lines-and-circles. At worst, this is just a computational thing, but/and the fact is important.

Then, since both the disk and the half-plane are bounded by a circle-or-line, we just have to track three points on the circle: one will go to infinity, so we know we'll get a line ("circle through infinity on the Riemann sphere"), so it only remains to compute the images of two other points from the circle... and they'll be on the imaginary axis. Yes, there is the subordinate issue of "which side of the imaginary axis", which can be tested by computing the image of a point from inside the circle, such as $0$.

Answer 3

In example 3 The following map is discussed: $$w=-{{z+1}\over{z-1}}$$ The author claims that this linear fractional transformation maps the unit disk onto the right halfplane but does not prove it. How it can be proved?

Alternative approach:

The problem is simple enough, that I see no need for elegance.

Let $z = x + iy$. Then, by the constraints, you know that

• $x^2 + y^2 \leq 1.$

• $(x,y) \neq (1,0)$.
  This follows, since otherwise the map of $w$ would be undefined at $z = 1$.

Note that the 2nd constraint above implies that $x < 1.$

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Then

$~\displaystyle w = - \frac{(x + 1) + iy}{(x - 1) + iy} \times \frac{(x - 1) - iy}{(x - 1) - iy} $

$~\displaystyle =~ - \frac{(x^2 - 1 + y^2) - i(2y)}{(x - 1)^2 + y^2}$

$$ = \frac{(1 - x^2 - y^2) + i(2y)}{(x-1)^2 + y^2}. \tag1 $$

In (1) above, since $x < 1$, the denominator must be a positive real number. In the numerator, the real component must be non-negative.

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It remains to consider what happens if $z = 1$ is allowed to be part of the domain.

If $z \to 1$, the denominator in (1) is approaching $0$ from above.

If $z \to 1$, with $z$ on the edge of the unit circle, then in (1), the real component is fixed at $(0)$.

If $z \to 1$, with $z$ not on the edge of the unit circle, then in (1), the real component is approaching $0$, from above.

Answer 4

By direct calculation (without assuming knowledge of Möbius transformations):

$$ \require{cancel} \begin{align} 2\, \text{Re}(w)= w+\bar w &= \frac{1+z}{1-z} + \frac{1+ \bar z}{1 - \bar z} \\ &= \frac{(1 - \cancel{\bar z} + \bcancel z - |z|^2) + (1 - \bcancel z + \cancel{\bar z} - |z|^2)}{(1-z)(1-\bar z)} \\ &= 2\,\frac{1-|z|^2}{|1-z|^2} \end{align} $$

Thus $\,|z| \lt 1 \iff \text{Re}(w) \gt 0\,$ i.e. the interior of the unit disk maps to the right half-plane.

Answer 5

Here is a geometric argument.

The imaginary axis is the bisector of the segment $[-1,1]$ and thus consists of points that are equidistant from $1$ and $-1$. The right half plane consists of points closer to $1$ and the left half plane consists of points closer to $-1$.

Note that $|z-a|$ is the distance of $z$ from $a$ in the plane.

Thus $|z-1|<|z+1|$ if and only if $z$ is closer to $1$ than it is to $-1$. As seen above this happens if and only if $z$ lies in the right half plane.

Update: It was pointed out that this is in the wrong direction. Let $w=(z-1)/(z+1)$. We have proved that $|w|<1$ if and only if $z$ lies in the right half-plane.

Note that by "componendo-dividendo" we have $$ \frac{w}{1} = \frac{z-1}{z+1} ~\text{if and only if}~ \frac{1+w}{1-w} = \frac{(z+1)+(z-1)}{(z+1)-(z-1)}=\frac{z}{1} $$ Thus, it is clear that $z=(1+w)/(1-w)$ maps the disk $|w|<1$ to the right half-plane $\mathrm{Re}(z)>0$.

Answer 6

We have $w = -\frac{2+(z-1)}{z-1}=-(1+\frac 2 {z-1})$. $\frac 2 {z-1}$ is a circle inversion, with the inversion circle being a radius-2 circle centered at $z=1$. The unit circle goes through the center of the inversion circle, so it's sent to a line. The unit circle is tangent to the inversion circle at $z=-1$. The inversion circle consists of fixed points, so the line the unit circle is sent to is also tangent to the inversion circle at $z=-1$, hence it is the line $Re(z)=-1$.

So that takes care of the $\frac 2 {z-1}$ part. Then we add $1$ and then multiply the result by $-1$. Adding $1$ moves the line to $Re(z)=0$. Since the unit disk is to the left of the inversion circle's center, the image after inversion is also to the left. When we multiply by $-1$, we flip that and get the right-side half plane.