Elementary mapping studio: powers, exponentials and Möbius maps

Terminology used in this mapping studio

At an interior point $a$ of a complex domain, the complex derivative is $f'(a)=\lim_{h\to0}(f(a+h)-f(a))/h$ when the same finite limit exists as h approaches zero along any complex path. A function is holomorphic on an open set when this derivative exists at every point there, and entire when it is holomorphic on all of $\mathbb C$. A holomorphic point where the derivative vanishes is a critical point. An antiholomorphic function is locally the complex conjugate of a holomorphic function. A zero is simple when the first derivative at that zero is nonzero.

1. Map a region by solving inequalities, not only its boundary

For $w=f(z)$, mapping a region requires determining all image points and their multiplicities, not just drawing images of a few boundary curves. A reliable workflow is: identify undefined points; find an inverse or parameter relation; translate the original domain conditions; check whether every proposed image point has an allowed preimage; then inspect orientation and critical points.

For $w=z^2$, polar coordinates give $|w|=|z|^2$ and an argument doubled modulo $2\pi$. On the sector $0<\arg z<\pi/2$, the map is one-to-one onto the upper half-plane: its argument lies in $(0,\pi)$ and its modulus is any positive number. Conversely every upper-half-plane point has a unique square root with angle in $(0,\pi/2)$. On the full punctured plane it is two-to-one, so local angle preservation where $2z\ne0$ does not imply global injectivity.

Worked example: a hyperbola becomes a line. Write $z=x+iy$ and $w=u+iv=z^2$. Then $u=x^2-y^2$ and $v=2xy$. The curves $x^2-y^2=c$ map into vertical lines $u=c$, while $2xy=d$ map into horizontal lines $v=d$. To claim an entire line as an image, one still checks branch restrictions on the original curve. The two families meet orthogonally away from zero; the origin is the critical point.

2. Inversion exchanges bounded and unbounded geometry

For $w=1/z$, $z\ne0$,

$$w=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}.$$

Thus $|w|=1/|z|$ and the argument changes sign. This does not make inversion antiholomorphic: its derivative is $-1/z^2\ne0$, so it preserves oriented angles locally. Reflection of polar angle is accompanied by reversal of radial direction; examining just one coordinate does not determine planar orientation.

The unit disk punctured at zero maps to $|w|>1$, and the upper half-plane maps to the lower half-plane because $\operatorname{Im}(1/z)=-y/|z|^2$. A circle passing through zero can map to a line, as the equation

$$|z-1|=1\quad\Longleftrightarrow\quad |z|^2-2\operatorname{Re}z=0$$

becomes $1-2\operatorname{Re}w=0$ after dividing by $|z|^2$. The point zero itself was excluded; the image is the entire line $\operatorname{Re}w=1/2$, as the inverse calculation verifies.

3. Möbius transformations and the half-plane–disk correspondence

A Möbius transformation has the form

$$w=\frac{az+b}{cz+d},\qquad ad-bc\ne0.$$

Where its denominator is nonzero it is holomorphic with derivative $(ad-bc)/(cz+d)^2\ne0$. Solving for the inverse gives

$$z=\frac{dw-b}{a-cw}.$$

It extends to a bijection of the extended plane $\widehat{\mathbb C}=\mathbb C\cup\{\infty\}$ by sending its pole to infinity and infinity to the appropriate limiting value. Without the determinant condition the expression can collapse to a constant, with a removable hole in its written formula; it is not then a Möbius bijection.

Generalized circles: nondegenerate conditions and a complete transformation proof

A genuine circle or line is not just any equation of the following shape. Put

$$A|z|^2+Bz+\bar B\bar z+C=0,\qquad A,C\in\mathbb R,\quad B\in\mathbb C,\quad D=|B|^2-AC.$$

If $A\ne0$ and $D>0$, completing the square gives the genuine circle

$$\left|z+\frac{\bar B}{A}\right|^2=\frac{D}{A^2}.$$

If $A=0$ and $B\ne0$, the equation is the genuine line $2\operatorname{Re}(Bz)+C=0$, and again $D>0$. Other coefficient choices may give an empty set, a point or a degenerate equation; they are not being called genuine generalized circles here.

To keep finite points and infinity under control, represent a point of the extended plane by a nonzero pair $(Z,T)$, where pairs differing by a nonzero complex factor represent the same point. A finite point is represented by $(z,1)$ and infinity by $(1,0)$. The homogeneous circle equation is

$$q(Z,T)=A|Z|^2+BZ\bar T+\bar B\bar ZT+C|T|^2=0.$$

Multiplying both coordinates by a nonzero factor multiplies this expression by that factor's squared modulus, so its zero set is well-defined. Infinity lies in it exactly when $A=0$. Thus a line includes infinity in this extended-plane interpretation, while a genuine circle does not.

Let $\Delta=ad-bc\ne0$. For the Möbius map in this section, an inverse homogeneous transformation is

$$\begin{pmatrix}Z\\T\end{pmatrix}=N\begin{pmatrix}W\\S\end{pmatrix},\qquad N=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.$$

The matrix $N$ is invertible because $\det N=\Delta$. Substituting this nonzero pair in the homogeneous equation is therefore reversible and proves equality of the full image set, not merely inclusion. At a finite image point set $(W,S)=(w,1)$. Expansion gives

$$A'|w|^2+B'w+\overline{B'}\bar w+C'=0,$$

where

$$A'=A|d|^2-Bd\bar c-\bar B\bar d c+C|c|^2,$$ $$B'=-Ad\bar b+Bd\bar a+\bar B\bar b c-Cc\bar a,$$ $$C'=A|b|^2-Bb\bar a-\bar B\bar b a+C|a|^2.$$

The first and last coefficients are real.

It remains to exclude a degenerate transformed set. Write

$$H=\begin{pmatrix}A&\bar B\\B&C\end{pmatrix},\qquad q(Z,T)=\begin{pmatrix}\bar Z&\bar T\end{pmatrix}H\begin{pmatrix}Z\\T\end{pmatrix}.$$

With $N^*$ denoting conjugate transpose, substitution gives the Hermitian matrix $H'=N^*HN$, whose entries are the displayed transformed coefficients. For two-by-two matrices the determinant product identity follows by expanding:

$$\det\left(\begin{pmatrix}p&q\\r&s\end{pmatrix}\begin{pmatrix}u&v\\w&x\end{pmatrix}\right)=(pu+qw)(rv+sx)-(pv+qx)(ru+sw)=(ps-qr)(ux-vw).$$

Also $\det(N^*)=\overline{\det N}$. Hence

$$A'C'-|B'|^2=\det H'=|\Delta|^2\det H=|\Delta|^2(AC-|B|^2),$$

so

$$|B'|^2-A'C'=|\Delta|^2D>0.$$

If $A'\ne0$, completing the square therefore gives a genuine circle with positive squared radius. If $A'=0$, the last inequality forces $B'\ne0$, so it gives a genuine line. This proves that every genuine generalized circle maps to a genuine generalized circle, including the circle-to-line case.

The homogeneous argument also prevents an incorrect finite-point conclusion after clearing a denominator. If the original set contains infinity, the transformed set includes its image; that image need not have a finite original preimage. If the original set contains the Möbius pole, its image is infinity. To obtain the image of only the original finite non-pole points, remove any transformed point that came solely from the omitted original infinity or pole. Such removals concern the finite-domain version of the map, not failure of the extended-plane theorem.

For example, inversion of the circle $|z-1|=1$ uses $A=1$, $B=-1$, $C=0$ and the map $w=1/z$. The formulas give $A'=0$, $B'=-1$, $C'=1$, hence $\operatorname{Re}w=1/2$, with positive discriminant. The original circle contains the pole zero, so the complete extended image is that line together with infinity; its finite non-pole image is precisely the finite line already found in the previous section.

Worked example. The Cayley map

$$w=\frac{z-i}{z+i}$$

sends the upper half-plane bijectively to the unit disk. Indeed,

$$|z+i|^2-|z-i|^2=4\operatorname{Im}z,$$

so $\operatorname{Im}z>0$ gives $|w|<1$. Its inverse is $z=i(1+w)/(1-w)$, for which

$$\operatorname{Im}z=\frac{1-|w|^2}{|1-w|^2}>0$$

when $|w|<1$. Thus both inclusion and surjectivity are proved. On the real axis its modulus is one, but the image of finite real points omits $w=1$; that boundary point corresponds to $z=\infty$.

4. Exponential strips and logarithmic coordinates

For $z=x+iy$, $e^z=e^xe^{iy}$. A horizontal line $y=\theta$ maps to the ray of argument $\theta$; a vertical line $x=c$ maps to the circle of radius $e^c$. On the strip $\alpha<\operatorname{Im}z<\beta$ with $0<\beta-\alpha\le2\pi$, exponential is one-to-one and maps onto the sector with a continuous argument between those bounds. When the width is $2\pi$, the image is a slit plane: the common boundary ray is omitted, not counted twice. If the width exceeds $2\pi$, points whose imaginary parts differ by $2\pi$ can both lie inside, so injectivity fails.

The principal logarithm is the inverse of exponential on $-\pi<\operatorname{Im}z<\pi$, mapping the slit plane to that strip. It turns multiplicative radial distance into an additive real coordinate. But $\operatorname{Log}(zw)=\operatorname{Log}z+\operatorname{Log}w$ fails when the selected arguments cross the branch cut, as C01 explains. Region mapping through a logarithm must specify which continuous argument is being used.

Worked example. The annular sector $1<|z|<e^2$, $0<\arg z<\pi/3$ maps under its chosen logarithm to the rectangle $0<\operatorname{Re}w<2$, $0<\operatorname{Im}w<\pi/3$. Exponentiating the rectangle supplies the inverse and proves the exact image.

5. Complex trigonometric functions are entire, but not bounded

Define

$$\sin z=\frac{e^{iz}-e^{-iz}}{2i},\qquad \cos z=\frac{e^{iz}+e^{-iz}}2.$$

The exponential derivative gives $\sin'=\cos$, $\cos'=-\sin$, and algebra gives $\sin^2z+\cos^2z=1$. For $z=x+iy$,

$$\sin z=\sin x\cosh y+i\cos x\sinh y,$$ $$\cos z=\cos x\cosh y-i\sin x\sinh y,$$

where $\cosh y=(e^y+e^{-y})/2$ and $\sinh y=(e^y-e^{-y})/2$. Thus $|\sin z|\le1$ is false for complex inputs: $\sin(iy)=i\sinh y$ is unbounded as $|y|\to\infty$.

The zeros of $\sin z$ satisfy $e^{2iz}=1$, so $z=k\pi$ for integers $k$; they are simple because $\cos(k\pi)=(-1)^k\ne0$. Similarly cosine vanishes at $\pi/2+k\pi$. The quotient $\tan z$ is holomorphic wherever cosine is nonzero, but it is not entire. Indeed, at $a=\pi/2+k\pi$ we have $\cos a=0$ and $\cos'(a)=-\sin a\ne0$. The derivative definition and continuity of sine give

$$\lim_{z\to a}(z-a)\tan z=\frac{\sin a}{\cos'(a)}=-1.$$

This finite nonzero limit is the simple-pole behavior used here: the function grows like a nonzero multiple of $1/(z-a)$. The later singularity chapters develop the general classification. These elementary exponential definitions already establish the holomorphy and derivative identities; series-representation results will justify the corresponding Taylor expansions.

6. Exercises

P01 — Foundation. Map the sector $0<\arg z<\pi/3$ under $z^3$, and prove injectivity there.

P02 — Core. Find the image of $|z|>2$ under $1/z$, distinguishing whether zero belongs to it.

P03 — Core. Determine the image of the upper half-plane under $(z-i)/(z+i)$ and verify its inverse.

P04 — Core. Find the image under exponential of $0<\operatorname{Re}z<1$, $0<\operatorname{Im}z<\pi/2$.

P05 — Diagnosis. Is the image of the finite real axis under the Cayley map the whole unit circle? Explain the missing point.

P06 — Proof. Show that a Möbius transformation has no critical point in its finite holomorphy domain.

P07 — Core. Evaluate $\sin(2i)$ and $\cos(2i)$ in terms of real hyperbolic functions. Verify the trigonometric identity there.

P08 — Extension. Find a Möbius map sending the open right half-plane to the unit disk, and prove both inclusion and surjectivity.

7. Full solutions

P01. Argument is tripled from $(0,\pi/3)$ to $(0,\pi)$ and positive modulus is cubed bijectively, so the image is the upper half-plane. Every image point has exactly one cube root with its argument in that interval, proving injectivity and surjectivity.

P02. The inequalities give $0<|w|<1/2$. Zero has no finite preimage under inversion; it would correspond to infinity on the extended plane. Thus the image is a punctured disk, not the full disk.

P03. The calculation in §3 gives $|w|<1$. Solving yields $z=i(1+w)/(1-w)$, whose imaginary part is $(1-|w|^2)/|1-w|^2>0$. This proves the exact unit-disk image, not just that the real boundary maps into the circle.

P04. It is the annular sector $1<|w|<e$, $0<\arg w<\pi/2$. The chosen argument gives a unique logarithmic preimage in the rectangle.

P05. No. Equality $(z-i)/(z+i)=1$ would imply $-i=i$, impossible for finite $z$. Every other unit-circle point has a finite real inverse by the displayed formula. The omitted point one is the image of infinity.

P06. Differentiation gives $(ad-bc)/(cz+d)^2$. Its numerator is nonzero by definition and its denominator is finite and nonzero in the holomorphy domain. Therefore the derivative cannot vanish there.

P07. $\sin(2i)=i\sinh2$ and $\cos(2i)=\cosh2$. Their squared sum is $-\sinh^22+\cosh^22=1$, using the exponential definitions. Their moduli need not lie below one.

P08. Take $w=(z-1)/(z+1)$. For $\operatorname{Re}z>0$, $|z+1|^2-|z-1|^2=4\operatorname{Re}z>0$, so $|w|<1$. The inverse $z=(1+w)/(1-w)$ has real part $(1-|w|^2)/|1-w|^2>0$, proving surjectivity. The pole at $-1$ lies outside the right half-plane.

8. Summary

Power maps multiply angles, inversion changes radial scale, Möbius maps exchange generalized circles, and logarithms turn annular sectors into rectangles. Exact image statements require an inverse or an equivalent surjectivity check. Complex trigonometric functions inherit analyticity from exponential, not the boundedness properties of their real restrictions.