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MTH 5103 COMPLEX VARIABLES
LECTURE NOTES, WEEK 2
2
Complex Functions
In this Lecture we shall discuss the basic properties of functions of one complex
variable.
2.1
Functions of a Complex Variable
Let S ⊂ C. A function f on S is a rule which assigns to each z ∈ S a complex
number f (z), called the value of f at z. Examples of (complex) functions include
Polynomials e.g., f (z) = z 2 or f (z) = 1 + iz − z 2 + (1 + i)z 3
z
.
z2 + 2
These are defined everywhere except at roots z of the denominator polynomial.
Rational Functions (One polynomial divided by another) e.g., f (z) =
The Exponential Function e.g., f (z) = ez , which is defined to be the sum
z
e =
∞
X
zn
n=0
n!
=1+
z
z2 z3
+
+
+ ··· ,
1! 2!
3!
(1)
or equally well, by Euler’s Formula, as ez = ex (cos y+i sin y). We shall see later
that the series above converges absolutely for all z ∈ C, therefore f (z) = ez
can be defined for every z ∈ C (i.e., we can choose S = C).
Trigonometric and Hyperbolic Functions For complex z, we define
cos z =
eiz + e−iz
,
2
sin z =
eiz − e−iz
,
2i
(2)
which agrees with the definition for z real in Euler’s Formula. By analogy with
the real case, we define
cosh z =
ez + e−z
,
2
1
sinh z =
ez − e−z
.
2
(3)
With these definitions, it is easy to check the following identity
cos z + i sin z = eiz = cosh(iz) + sinh(iz).
(4)
Now, given any complex number may be expressed in the form x + iy, we may
write a complex function f as
f (z) = <(f (z)) + i=(f (z)) = u(x, y) + iv(x, y).
(5)
In this manner, we may think of f as a (real) mapping
f : R2 −→ R2
(x, y) 7−→ (u(x, y), v(x, y)) .
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Example 1. Rewrite the following complex functions as real mappings: z 2 , , ez .
z
1. f (z) = z 2 can be written as f (x, y) = (x2 − y 2 , 2xy).
1
x
−y
2. f (z) = can be written as f (x, y) =
,
.
z
x2 + y 2 x2 + y 2
3. f (z) = ez = ex+iy can be written as f (x, y) = (ex cos y, ex sin y).
2.2
Transformations of the Complex Plane
A complex function is a map (transformation) C → C, but to draw a graph of such a
mapping, we would need 2+2=4 real dimensions. So instead, we shall examine what
happens to various “shapes” (e.g., lines, curves, etc.) under such a map.
Example 2. Examine the mapping w = z 2 . In the previous example, we wrote
down this mapping in (x, y), (u, v) coordinates as u + iv = x2 − y 2 + 2ixy, i.e.,
u = x2 − y 2 and v = 2xy. It is easier, however, to understand this map if we write
z and w in polar coordinates. Then for z = reiθ , we have f (z) = w = r2 e2iθ .
We first determine where the following curves are mapped under f :
The real axis R is mapped to the positive real axis R+ : We can see this by
representing the real axis in two pieces, the positive real axis by points of
the form rei0 = r > 0 and the negative real axis by reiπ = −r < 0. These
points are mapped under f to r2 ei0 = r2 ei2π = r2 , which lies on the positive
real axis.
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The imaginary axis iR is mapped to the negative real axis R− : Using a
similar argument as above, we can represent the positive imaginary axis
using points of the form reiπ/2 = ir and those on the negative imaginary
axis as rei3π/2 = −ir (r > 0). These points are mapped under f to r2 eiπ =
r2 ei3π = −r, which lies on the negative real axis.
A quarter circle of radius r centred at 0 maps to a half circle of radius r2
centred at 0: The quarter circle is given byreiθ where 0 ≤ θ ≤ π/2, which
is mapped to r2 e2iθ . Notice that as θ ranges from 0 to π/2, the angle 2θ
ranges from 0 to π, giving us the half circle.
A line `θ passing through the origin at angle θ maps to a ray `+
2θ starting at
0 making an angle 2θ: In this case, the line is parametrised by reiθ , where
θ is now fixed, but r may vary. The mapping sends r to r2 , resulting in a
ray (and the angle is transformed from θ to 2θ).
Example 3. What is the image of a vertical line x = c ∈ R under the mapping w =
z 2 ? Recall that the function z 7→ z 2 can be expressed as (u, v) = (x2 − y 2 , 2xy).
Setting x = c, we have the system
u = c2 − y 2 ,
v = 2cy,
from which we may eliminate the variable y to obtain the equation for a curve in
v2
the u − v plane (equaivalently, the w-plane). Doing this gives us u = c2 − 2 ,
4c
which is a leftward-opening parabola with vertex (c2 , 0), intersecting the v axis
at (0, ±2c2 ).
Example 4. Using the results of the previous examples, determine which regions,
if any, of the z-plane are mapped to the upper half of the w plane under the
mapping w = z 2 .
First, we identify the upper half of the w-plane to be {w ∈ C : =(w) ≥ 0}.
An element z = reiθ maps to w = r2 e2iθ in the upper half plane if and only if the
argument of w lies between 0 and π. That is,
0 + 2kπ ≤ 2θ ≤ π + 2kπ
π
kπ ≤ θ ≤ kπ +
2
(k ∈ Z)
(k ∈ Z),
which means that z must lie in the first or third quadrant in order to map to the
upper half plane (plug in k = 0, 1, 2, 3, . . . to see this).
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Alternatively, we could use the (x, y) and (u, v) coordinates to solve this
problem as follows: w lies in the upper half plane if and only if v = 2xy ≥ 0. But
this is true only when x and y are either both ≥ 0 or both ≤ 0, that is, when z
lies in the first or third quadrant.
Proposition 1. The following kinds of transformations all send straight lines to
straight lines and circles to circles:
1. Translations w = z + c, for complex constant c
2. Rotations w = eiθ z for a real constant θ
3. Dilations w = rz for a real, nonzero constant r
4. Linear or Affine Maps w = λz + c for complex constants λ, c with λ 6= 0.
5. Complex Conjugation w = z̄
Proof. Translations, rotations and dilations all obviously satisfy that they map
lines to lines and circles to circles (convince yourself geometrically!). To prove
linear mappings also satisfy this property, we rewrite λ = reiθ and observe that
λz +c can be written as a composition of a dilation followed by a rotation followed
by a translation, reading left to right:
z 7−→ rz 7−→ reiθ z 7−→ reiθ z + c.
(6)
Finally, note that complex conjugation z = (x, y) 7−→ z̄ = (x, −y) is simply
reflection about the x-axis which also sends lines to lines and circles to circles.
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Another function we often need to consider is z 7−→ w = . We may also regard
z
this as a composition (using the fact that z z̄ = |z|2 ):
z 7−→
z̄
1
z
7−→ 2 = .
2
|z|
|z|
z
(7)
The first mapping is “inversion” about the unit circle and the second is complex conjugation; it is an instructive exercise to draw a picture of each of these mappings and
determine where various regions of the plane are mapped (e.g., in the first quadrant,
the interior of unit circle maps to the exterior of unit circle in the fourth quadrant).
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Consequently we can write down how z 7−→ w = maps the punctured plane C\{0}
z
to itself (or C ∪ {∞} to itself).
4
1
maps
z
1. circles not through 0 to circles not through 0,
Proposition 2. z 7−→ w =
2. circles through 0 to straight lines not through 0,
3. straight lines not through 0 to circles through 0,
4. straight lines through 0 to straight lines through 0.
Remark 1. If we regard a straight line as a “circle passing through ∞” then the
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above Proposition says the mapping z 7−→ sends circles to circles.
z
Proof. We outline the main ideas of the proof. Consider the equation
a(x2 + y 2 ) + bx + cy + d = 0
(a, b, c, d ∈ R).
(8)
If a = 0, this is a straight line. If a 6= 0, then by completing
the squares in x
−b −c
,
and y separately, we see this is a circle centered at
. In either case, it
2a 2a
passes through 0 ∈ C (equivalently, (0, 0) ∈ R2 ) if and only if d = 0.
1
Now let w = so that for z = x + iy and w = u + iv,
z
1
z =
w
v
u
−i 2
x + iy = 2
2
u +v
u + v2
u
−v
⇒x= 2
,
y= 2
.
2
u +v
u + v2
Substituting for x and y in (8) and clearing denominators, we obtain
d(u2 + v 2 ) + bu − cv + a = 0.
(9)
But comparing this equation directly with (8), we see it must be a circle in the
w-plane (equivalently, the (u, v)-plane) if and only if d 6= 0 and it is a line if and
only if d = 0. Furthermore, it passes through the origin if and only if a = 0,
which proves the Proposition.
Example 5. Determine how the (horizontal) line y = k is transformed under
1
u
−v
z 7→ . In this case, we may substitute x = 2
,y = 2
into the equa2
z
u +v
u + v2
tion y = k to find
k(u2 + v 2 ) + v = 0.
(10)
1
This is simply a circle in the w-plane centered at 0, −1
of radius 2k
.
2k
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2.3
Möbius Transformations
Let a, b, c, d ∈ C be such that ad − bc 6= 0. A mapping of the form
z 7−→ w =
az + b
cz + d
(11)
is called a fractional linear map or a Möbius transformation. Visibly, any Möbius
transformation can be expressed as a composition of a linear map with an inversion,
followed by another linear map:
1
1
7−→ α
+ β = w.
(12)
z 7−→ cz + d 7−→
cz + d
cz + d
Notice that if we require w =
az + b
, the solving for α, β gives us
cz + d
α + β(cz + d) = az + b
⇒
β=
a
ad
and α = b − .
c
c
Given the above calculation, we have immediately from Propositions 1 and 2
Proposition 3. Fractional linear transformations (Möbius transformations) send
lines and circles to lines and circles.
Möbius transformations are often represented by their corresponding!matrices,
a b
az + b
e.g., for z 7→
, the associated matrix is given by M =
. It is a
cz + d
c d
straightforward calculation (do this!) to verify that the composition of two Möbius
transformations is itself a Möbius transformation and that the matrix associated with
the resulting transformation can be found by multiplying the corresponding matrices
of the composition transformations.
az + b
Remark 2. z 7−→
can be extended to a map from C ∪ {∞} to C ∪ {∞} by
cz + d
a
−d
sending z = ∞ to w = and sending z =
to w = ∞. It is a bijection from
c
c
C ∪ {∞} to itself and has inverse given by
w 7−→ w =
dw − b
.
−cw + a
(13)
Example 6. Find the Möbius transformation sending −1 7→ i, 0 7→ 1, and 1 7→ −i.
To do this, we require
i=
−a + b
,
−c + d
b
1= ,
d
6
−i =
a+b
.
c+d
(14)
From the second equation we have d = b so we may eliminate the other variables
in the remaining equations to find a = −ib and c = ib. Hence the Möbius
transformation is given by
z 7−→
−ibz + b
z+i
=
.
ibz + b
−z + i
(15)
Note that under this transformation, the real axis is mapped to the unit circle
(R+ 7→ lower semicircle and R− 7→ upper semicircle), and the upper-half of the
z-plane is mapped the exterior of this circle (first quadrant 7→ lower exterior and
second quadrant 7→ upper exterior).
It is a harder exercise to show that given any three distinct points z1 , z2 , z3 in
the z-plane, and any three points w1 , w2 , w3 in the w-plane, there exists a Möbius
transformation sending zj 7→ wj for each j = 1, 2, 3. This transformation is unique if
we specify an orientation for the curve containing the three given points (or for the
curve containing the prescribed points, but not both).
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