3.1: Basic Transformations of Complex Numbers
( \newcommand{\kernel}{\mathrm{null}\,}\)
We begin with a definition.
Given two sets A and B, a function f:A→B is called one-to-one (or 1-1) if whenever a1≠a2 in A, then f(a1)≠f(a2) in B. The function f is called onto if for any b in B there exists an element a in A such that f(a)=b. A transformation on a set A is a function T:A→A that is one-to-one and onto.
Following are two schematics of functions. In the first case, f:A→B is onto, but not one-to-one. In the second case, g:A→B is one-to-one, but not onto.
A transformation T of A has an inverse function, T−1, characterized by the property that the compositions T−1∘T(a)=a and T∘T−1(a)=a for all a in A. The inverse function T−1 is itself a transformation of A and it “undoes” T in this sense: For elements z and w in A, T−1(w)=z if and only if T(z)=w.
In this section, we develop the following basic transformations of the plane, as well as some of their important features.
- General linear transformation: T(z)=az+b, where a,b are in C with a≠0.
- Special cases of general linear transformations:
- Translation by b: Tb(z)=z+b.
- Rotation by θ about 0: Rθ(z)=eiθz.
- Rotation by θ about z0: R(z)=eiθ(z−z0)+z0.
- Dilation by factor k>0: T(z)=kz.
- Reflection across a line L: rL(z)=eiθ¯z+b, where b is in C, and θ is in R.
Consider the fixed complex number b, and define the function Tb:C→C by
Tb(z)=z+b.
The notation helps us remember that z is the variable, and b is a complex constant. We will prove that Tb is a transformation, but this fact can also be understood by visualizing the function. Each point in the plane gets moved by the vector b, as suggested in the following diagram.
For instance, the origin gets moved to the point b, (i.e., Tb(0)=b), and every other point in the plane gets moved the same amount and in the same direction. It follows that two different points, such as v and w in the diagram, cannot get moved to the same image point (thus, the function is one-to-one). Also, any point in the plane is the image of some other point (just follow the vector −b to find this “pre-image” point), so the function is onto as well.
We now offer a formal argument that the translation Tb is a transformation. Recall, b is a fixed complex number.
That Tb is onto:
To show that Tb is onto, let w denote an arbitrary element of C. We must find a complex number z such that Tb(z)=w. Let z=w−b. Then Tb(z)=z+b=(w−b)+b=w. Thus, T is onto.
That Tb is one-to-one:
To show that Tb is 1-1 we must show that if z1≠z2 then Tb(z1)≠Tb(z2). We do so by proving the contrapositive. Recall, the contrapositive of a statement of the form “If P is true then Q is true” is “If Q is false then P is false.” These statements are logically equivalent, which means we may prove one by proving the other. So, in the present case, the contrapositive of “If z1≠z2 then Tb(z1)≠Tb(z2)” is “If Tb(z1)=Tb(z2), then z1=z2.” We now prove this statement.
Suppose z1 and z2 are two complex numbers such that Tb(z1)=Tb(z2). Then z1+b=z2+b. Subtracting b from both sides we see that z1=z2, and this completes the proof.
Let θ be an angle, and define Rθ:C→C by Rθ(z)=eiθz.
This transformation causes points in the plane to rotate about the origin by the angle θ. (If θ>0 the rotation is counterclockwise, and if θ<0 the rotation is clockwise.) To see this is the case, suppose z=reiβ, and notice that
Rθ(z)=eiθreiβ=rei(θ+β).
To achieve a rotation by angle θ about a general point z0, send points in the plane on a three-leg journey: First, translate the plane so that the center of rotation, z0, goes to the origin. The translation that does the trick is T−z0. Then rotate each point by θ about the origin (Rθ). Then translate every point back (Tz0). This sequence of transformations has the desired effect and can be tracked as follows:
zT−z0⟼z−z0Rθ⟼eiθ(z−z0)Tz0⟼eiθ(z−z0)+z0.
In other words, the desired rotation R is the composition Tz0∘Rθ∘T−z0 and
R(z)=eiθ(z−z0)+z0.
That the composition of these three transformations is itself a transformation follows from the next theorem.
If T and S are two transformations of the set A, then the composition S∘T is also a transformation of the set A.
- Proof
-
We must prove that S∘T:A→A is 1-1 and onto.
That S∘T is onto:
Suppose c is in A. We must find an element a in A such that S∘T(a)=c.
Since S is onto, there exists some element b in A such that S(b)=c.
Since T is onto, there exists some element a in A such that T(a)=b.
Then S∘T(a)=S(b)=c, and we have demonstrated that S∘T is onto.
That S∘T is 1-1:
Again, we prove the contrapositive. In particular, we show that if S∘T(a1)=S∘T(a2) then a1=a2.
If S(T(a1))=S(T(a2)) then T(a1)=T(a2) since S is 1-1.
And T(a1)=T(a2) implies that a1 = a2 since T is 1-1.
Therefore, S∘T is 1-1.
Suppose k>0 is a real number. The transformation T(z)=kz is called a dilation; such a map either stretches or shrinks points in the plane along rays emanating from the origin, depending on the value of k.
Indeed, if z=x+yi, then T(z)=kx+kyi, and z and T(z) are on the same line through the origin. If k>1 then T stretches points away from the origin. If 0<k<1, then T shrinks points toward the origin. In either case, such a map is called a dilation.
Given complex constants a,b with a≠0 the map T(z)=az+b is called a general linear tranformation. We show in the following example that such a map is indeed a transformation of C.
Consider the general linear transformation T(z)=az+b, where a,b are in C and a≠0. We show T is a transformation of C.
That T is onto:
Let w denote an arbitrary element of C. We must find a complex number z such that T(z)=w. To find this z, we solve w=az+b for z. So, z=1a(w−b) should work (since a≠0, z is a complex number). Indeed, T(1a(w−b))=a⋅[1a(w−b)]+b=w. Thus, T is onto.
That T is one-to-one:
To show that T is 1-1 we show that if T(z1)=T(z2), then z1=z2.
If z1 and z2 are two complex numbers such that T(z1)=T(z2), then az1+b=az2+b. By subtracting b from both sides we see that az1=az2, and then dividing both sides by a (which we can do since a≠0), we see that z1=z2. Thus, T is 1-1 as well as onto, and we have proved T is a transformation.
Note that dilations, rotations, and translations are all special types of general linear transformations.
We will often need to figure out how a transformation moves a collection of points such as a triangle or a disk. As such, it is useful to introduce the following notation, which uses the standard convention in set theory that a∈A means the element a is a member of the set A.
Suppose T:A→A is a transformation and D is a subset of A. The image of D, denoted T(D), consists of all points T(x) such that x∈D. In other words,
T(D)={a∈A | a=T(x) for some x∈D}.
For instance, if L is a line and Tb is translation by b, then it is reasonable to expect that T(L) is also a line. If one translates a line in the plane, it ought to keep its linear shape. In fact, lines are preserved under any general linear transformation, as are circles.
Suppose T is a general linear transformation.
- T maps lines to lines.
- T maps circles to circles. Proof.
- Proof
-
- We prove that if L is a line in C then so is T(L). A line L is described by the line equation
αz+¯α¯z+d=0
for some complex constant α and real number d. Suppose T(z)=az+b is a general linear transformation (so a≠0). All the points in T(L) have the form w=az+b where z satisfies the preceeding line equation. It follows that z=1a(w−b) and when we plug this into the line equation we see that
αw−ba+¯α¯w−b¯a+d=0
which can be rewritten
αaw+¯α¯a¯w+d−αba−¯αb¯a=0.
Now, for any complex number β the sum β+¯β is a real number, so in the above expression, d−(αba+¯αb¯a) is a real number. Therefore, all w in T(L) satisfy a line equation. That is, T(L) is a line.
- The proof of this part is left as an exercise.
The image of the disk D={z∈C | |z−2i|≤1} under the transformation T:C→C given by T(z)=2z+(4−i) is the disk T(D) centered at 4+3i with radius 2 as pictured below.
We will be interested in working with transformations that preserve angles between smooth curves. A planar curve is a function r:[a,b]→C mapping an interval of real numbers into the plane. A curve is smooth if its derivative exists and is nonzero at every point. Suppose r1 and r2 are two smooth curves in C that intersect at a point. The angle between the curves measured from r1 to r2, which we denote by ∠(r1,r2), is defined to be the angle between the tangent lines at the point of intersection.
A transformation T of C preserves angles at point z0 if ∠(r1,r2)=∠(T(r1),T(r2)) for all smooth curves r1 and r2 that intersect at z0. A transformation T of C preserves angles if it preserves angles at all points in C. A transformation T of C preserves angle magnitudes if, at any point in C, |∠(r1,r2)|=|∠(T(r1),T(r2))| for all smooth curves r1 and r2 intersecting at the point.
General linear transformations preserve angles.
- Proof
-
Suppose T(z)=az+b where a≠0. Since the angle between curves is defined to be the angle between their tangent lines, it is sufficient to check that the angle between two lines is preserved. Suppose L1 and L2 intersect at z0, and zi is on Li for i=1,2, as in the following diagram.
Then,
∠(L1,L2)=arg(z2−z0z1−z0).
Since general linear transformations preserve lines, T(Li) is the line through T(z0) and T(zi) for i=1,2 and it follows that
∠(T(L1),T(L2))=arg(T(z2)−T(z0)T(z1)−T(z0))=arg(az2+b−az0−baz1+b−az0−b)=arg(z2−z0z1−z0)=∠(L1,L2).
Thus T preserves angles.
A fixed point of a transformation T:A→A is an element a in the set A such that T(a)=a.
If b≠0, the translation Tb of C has no fixed points. Rotations of C and dilations of C have a single fixed point, and the general linear transformation T(z)=az+b has one fixed point as long as a≠1. To find this fixed point, solve
z=az+b
for z. For instance, the fixed point of the transformation T(z)=2z+(4−i) of Example 3.1.6 is found by solving z=2z+4−i, for z, which yields z=−4+i. So, while the map T(z)=2z+(4−i) moves the disk D in the example to the disk T(D), the point −4+i happily stays where it is.
A Euclidean isometry is a transformation T of C with the feature that |T(z)−T(w)|=|z−w| for any points z and w in C. That is, a Euclidean isometry preserves the Euclidean distance between any two points.
It is perhaps clear that translations, which move each point in the plane by the same amount in the same direction, ought to be isometries. Rotations are also isometries. In fact, the general linear transformation T(z)=az+b will be a Euclidean isometry so long as |a|=1:
|T(z)−T(w)|=|az+b−(aw+b)|=|a(z−w)|=|a||z−w|.
So, |T(z)−T(w)|=|z−w|⟺|a|=1. Translations and rotations about a point in C are general linear transformations of this type, so they are also Euclidean isometries.
Reflection about a line L is the transformation of C defined as follows: Each point on L gets sent to itself, and if z is not on L, it gets sent to the point z∗ such that line L is the perpendicular bisector of segment zz∗.
Reflection about L is defined algebraically as follows. If L happens to be the real axis then
rL(z)=¯z.
For any other line L we may arrive at a formula for reflection by rotating and/or translating the line to the real axis, then taking the conjugate, and then reversing the rotation and/or translation.
For instance, to describe reflection about the line y=x+5, we may translate vertically by −5i, rotate by −π4, reflect about the real axis, rotate by π4, and finally translate by 5i to get the composition
z↦z−5i↦e−π4i(z−5i)↦¯e−π4i(z−5i)=eπ4i(¯z+5i)↦eπ4i⋅eπ4i(¯z+5i)=eπ2i(¯z+5i)↦eπ2i(¯z+5i)+5i.
Simplifying (and noting that eπ2i=i), the reflection about the line L:y=x+5 has formula
rL(z)=i¯z−5+5i.
In general, reflection across any line L in C will have the form
rL(z)=eiθ¯z+b
for some angle θ and some complex constant b.
Reflections are more basic transformations than rotations and translations in that the latter are simply careful compositions of reflections.
A translation of C is the composition of reflections about two parallel lines. A rotation of C about a point z0 is the composition of reflections about two lines that intersect at z0.
- Proof
-
Given the translation Tb(z)=z+b let L1 be the line through the origin that is perpendicular to segment 0b as pictured in Figure 3.1.18(a). Let L2 be the line parallel to L1 through the midpoint of segment 0b. Also let ri denote reflection about line Li for i=1,2.
Now, given any z in C, let L be the line through z that is parallel to vector b (and hence perpendicular to L1 and L2). The image of z under the composition r2∘r1 will be on this line. To find the exact location, let z1 be the intersection of L1 and L, and z2 the intersection of L2 and L, (see the figure). To reflect z about L1 we need to translate it along L twice by the vector z1−z. Thus r1(z)=z+2(z1−z)=2z1−z.
Next, to reflect r1(z) about L2, we need to translate it along L twice by the vector z2−r1(z). Thus,
r2(r1(z))=r1(z)+2(z2−r1(z))=2z2−r1(z)=2z2−2z1+z.
Notice from Figure 3.1.1(a) that z2−z1 is equal to b/2. Thus r2(r1(z))=z+b is translation by b.
Rotation about the point z0 by angle θ can be achieved by two reflections. The first reflection is about the line L1 through z0 parallel to the real axis, and the second reflection is about the line L2 that intersects L1 at z0 at an angle of θ/2, as in Figure 3.1.1(b). In the exercises you will prove that this composition of reflections does indeed give the desired rotation.
We list some elementary features of reflections in the following theorem. We do not prove them here but encourage you to work through the details. We will focus our efforts in the following section on proving analogous features for inversion transformations, which are reflections about circles.
Reflection across a line is a Euclidean isometry. Moreover, any reflection sends lines to lines, sends circles to circles, and preserves angle magnitudes.
In fact, one can show that any Euclidean isometry can be expressed as the composition of at most three reflections. See, for instance, Stillwell [10] for a proof of this fact.
Any Euclidean isometry is the composition of, at most, three reflections.
Exercises
Is T(z)=−z a translation, dilation, rotation, or none of the above?
Show that the general linear transformation T(z)=az+b, where a and b are complex constants, is the composition of a rotation, followed by a dilation, followed by a translation.
- Hint
-
View the complex constant a in polar form.
Prove that a general linear transformation maps circles to circles.
Suppose T is a rotation by 30∘ about the point 2, and S is a rotation by 45∘ about the point 4. What is T∘S? Can you describe this transformation geometrically?
Suppose T(z)=iz+3 and S(z)=−iz+2. Find T∘S. What type of transformation is this?
Find a formula for a transformation of C that maps the open disk D={z | |z|<2} to the open disk D′={z | |z−i|<5}. Is this transformation unique, or can you think of two different ones that work?
Find a formula for reflection about the vertical line x=k.
Find a formula for reflection about the horizontal line y=k.
Find a formula for reflection in the plane about the line y=mx+b, where m≠0.
- Hint
-
Think about what angle this line makes with the positive x-axis.
Prove that the construction in Figure 3.1.1(b) determines the desired rotation.
S(z)=kz is a dilation about the origin. Find an equation for a dilation of C by factor k about an arbitrary point z0 in C.