Lecture 7 : Complex Analysis
By Fu-Jiun Jiang
I.
INTRODUCTION
A complex number z can be written as z = a + ib, where a, b are
any 2 real numbers and are called the real and imaginary part of z,
respectively. Further the i appearing above is the square root of −1,
namely i2 = −1. The magnitude, argument and complex conjugate
of a complex number z = a + ib which are denoted by |z|, arg(z)
and z̄ are defined by
• |z| =
√
a2 + b2 .
• arg(z) = tan−1(b/a).
• z̄ = a − ib.
Notice a complex number z can be written as z = |z| exp (i arg(z)) as
well. The addition, subtraction, multiplication, division are defined
in a straighforward way. Let z1 = a1 + ib1 and z2 = a2 + ib2 be 2
complex numbers and r be a real number, then one has
• z1 + z2 = (a1 + ib1) + (a2 + ib2) = (a1 + a2) + i(b1 + b2).
• z1 − z2 = (a1 + ib1) − (a2 + ib2) = (a1 − a2) + i(b1 − b2).
1
• z1 × z2 = (a1 + ib1) × (a2 + ib2) = a1a2 − b1b2 + i(a1b2 + a2b1).
• z1/z2 = (a1 +ib1)/(a2 +ib2) = (a1a2 +b1b2)/(a22 +b22)+i(−a1b2 +
a2b1)/(a22 + b22).
• rz1 = r(c1 + ic2) = rc1 + irc2.
II.
FUNCITONS OF COMPLEX VARIABLE
One can define functions with complex variable. For example,
f (z) = az 2 + bz, here a, b are either real or complex numbers. One
also has exponential, trigonometric and other well-known functions
of complex variable :
• cos z = (exp(iz) + exp(−iz)) /2,
sin z = (exp(iz) − exp(−iz)) /(2i)...,
• exp z = exp(x + iy) = exp(x) × exp(iy)
= exp(x) × (cos y + i sin y),
• cosh z = (exp(z) + exp(−z)) /2,
sinh z = (exp(z) − exp(−z)) /2
Notice for the exponential function exp, one has exp(z + 2πi) =
exp(z). Further for the log function, one finds log (exp(z + 2πi)) =
z + 2πi. On the other hand, one also gets log (exp(z + 2πi)) =
2
log (exp(z)) = z. Hence we usually define the principal branch of
the logarithm function as
log z = log |z| + i arg(z).
III.
(1)
DIFFERENTIATION OF COMPLEX FUNCTIONS
The limit of a complex function at z0 = x0 + iy0 is similar to the
case of real functions. The definition of the limit of a function f at
a point z = z0 is essentially the same as that which we learned in
elementary calculus :
lim f (z) = L
(2)
z→z0
means that given an > 0, there is a δ so that |f (z) − L| <
whenever 0 < |z − z0| < δ. As you could guess, we say that
f is continuous at z0 if it is true that limz→z0 f (z) = f (z0). If f
is continuous at each point of its domain, we say simply that f is
continuous. Suppose both limz→z0 f (z) and limz→z0 g(z) exist, then
one has
lim (f (z) ± g(z)) = lim f (z) ± lim f (z),
z→z0
z→z0
z→z0
lim (f (z)g(z)) = lim f (z) lim f (z),
z→z0
z→z0
z→z0
lim (f (z)/g(z)) = lim f (z)/ lim g(z),
z→z0
z→z0
z→z0
3
g(z0) 6= 0.
(3)
Suppose f is a function and z0 is an interior point of the domain
of f . The derivative f 0(z0) of f is defined by
f (z) − f (z0)
.
z→z0
z − z0
f 0(z0) = lim
(4)
Now let’s see 2 example. Let f (z) = z 2 and ∆z = z − z0, then we
have
f (z0 + ∆z) − f (z0)
z→z0
∆z
2z0∆z − (∆z)2
= lim
∆→0
∆z
= lim (2z0 − ∆z)
lim = lim
z→z0
∆z→0
= 2z0.
(5)
Let f (z) = z z̄, then
f (z0 + ∆z) − f (z0)
z→z0
∆z
(z0 + ∆z)(z0 + ∆z) − z0z¯0
= lim
∆z→0
∆z
∆z
= lim (z0 + ∆z + z0 )
∆z→0
∆z
lim = lim
z→z0
(6)
Now let’s choose ∆z = ∆x, then we have
lim = z0 + z0
∆z→0
4
(7)
On the other hand, if we choose ∆z = i∆y, we find
lim = z0 − z0
∆z→0
(8)
In other word, f (z) = z z̄ does not have an unique derivative at
most of the complex plane, hence it is not differentiable at most of
the points on the complex plane.
If f has a derivative at z0, we say that f is differentiable at z0.
If f is differentiable at every point of a neighborhood of z0, we say
that f is analytic at z0. If f is analytic at every point of some set S,
we say that f is analytic on S. A function that is analytic on the set
of all complex numbers is said to be an entire function.
Suppose the function f given by f (z) = u(x, y) + iv(x, y) has a
derivative at z = z0 = x0 + iy0. We know this means there is a
number f 0(z0) so that
f (z0 + ∆z) − f (z0)
.
∆z→0
∆z
f 0(z0) = lim
(9)
Now let’s choose ∆z = ∆x, then we have
f 0(z0) =
∂u
∂v
(x0, y0) + i (x0, y0).
∂x
∂x
5
(10)
Next, if we choose ∆z = i∆y, we find
f 0(z0) =
∂v
∂u
(x0, y0) − i (x0, y0).
∂y
∂y
(11)
From above 2 equations, we immediately have
∂v
∂u
(x0, y0) =
(x0, y0)
∂x
∂y
∂u
∂v
(x0, y0) = − (x0, y0),
∂y
∂x
(12)
which are called Cauchy-Riemann equations.
We have shown that if f has a derivative at a point z0, then its
real and imaginary parts satisfy these equations. Even more exciting is the fact that if the real and imaginary parts of f satisfy
these equations and if in addition, they have continuous first partial
derivatives, then the function f has a derivative. Specifically, suppose u(x, y) and v(x, y) have partial derivatives in a neighborhood
of z0 = (x0, y0), suppose these derivatives are continuous at z0, and
suppose
∂v
∂u
(x0, y0) =
(x0, y0)
∂x
∂y
∂u
∂v
(x0, y0) = − (x0, y0),
∂y
∂x
6
(13)
then f is differentiable at z0.
IV.
COMPLEX INTEGRATION
Line, surface and volume integration over complex functions is
defined (and evaluated) similar to the integration over real functions.
For example, let C be the curve y =
1
x2
from the point z = 1 + i to
the point z = 3 + i 19 , then we have
1
i
260
728
z 2dz =
(3 + )3 − (1 + i)3 = −
−
i.
(14)
3
9
27
2187
C
R
Next let’se valuate the integral C z̄dz, where C is the parabola
Z
y = x2 from 0 to 1 + i :
Z
Z
(x − iy)(dx + idy)
z̄dz =
C
ZCx=1
=
Zx=0
1
=
(x − ix2)(dx + i2xdx)
x + 2x3 + ix2 dx
0
= .....
(15)
Now let’s turn to a very important theorem in complex analysis.
Let C1 and C2 be 2 curves in the complex plane with the same
end points. Further, if there exists a comtinuous function H(t, x, y)
7
so that H(0, x, y) = C1 and H(1, x, y) = C2. Then for any function
f which is analysis in the region bounded by C1 and C2, one has
Z
Z
f (z)dz =
C1
f (z)dz.
(16)
C2
This is called the Cauchy Theorem. Another statement of Cauchy
Theorem is as follows
Let f be an analysis function in a simple connected region D,
then for any closed curve which is within D, the one has
Z
f (z)dz = 0.
(17)
C
Proof : Let f (x) = u(x, y) + iv(x, y). then we have
Z
Z
(u(x, y) + iv(x, y))(dx + idy)
Z
Z
Z
Z
= ( udx − vdy) + i( udy + vdx). (18)
f (z)dz =
C(x,y)
C
Next, by Stoke theorem, we find
Z
Z
Z
∂v ∂u
− )dxdy
∂x
∂y
Z
Z
Z
∂u ∂v
udy + vdx = ( − )dxdy.
∂x ∂y
udx −
Hence
R
C
udy =
(−
(19)
f (z)dz = 0 since f (z) is analytic (so the real and imaginary
8
part of f (z) satisfy the Cauchy-Riemann equations).
Several important consequence of Cauchy theorem
• Let C be a simple closed curve bounding a region having a as
interior point, then one has
I
dz
= 2πi,
z
−
a
C
I
dz
= 0, n ≥ 2
n
(z
−
a)
IC
(z − a)ndz = 0, n ≥ 1.
(20)
C
Proof : Let C1 be a circle around a (a is the center of the circle)
R
dz
of radius . Then by Cauchy theorem, we have C (z−a)
n =
R
dz
C1 (z−a)n . Now by employing polar coodinate, for n = 1 we
arrive at
Z
dz
=
C1 z − a
Z
0
2π
ieiθ
dθ iθ = 2πi.
e
(21)
Similarly, one can easily show for n > 1 or n ≤ −1, one would
R
dz
have C1 (z−a)
n = 0.
• Let f (z) be analytic within and on a simple closed curve C and
9
a be any point interior to C , then
n!
f n(a) =
2πi
I
C
f (z)
dz.
(z − a)n+1
(22)
Let’s prove the case for n = 0 and the cases of general n ≥ 1 can
be proved by induction . Again let C1 be a circle around a (a is
the center of the circle) of radius . Then by Cauchy theorem,
R (z)dz
R (z)dz
R 2π
we have C f z−a
= C1 f z−a
= 0 dθif (a + eiθ ) = 2πif (a)
when → 0.
Let’s see 2 applications of Cauchy theorem. First of all we want
H cos z
to calculate the integral C z−π , where C is the circle |z − 1| = 3.
Notice since z = π is within the circle |z − 1| = 3, we see
I
C
cos z
= 2πi × cos π = −2πi.
z−π
Next we want to compute the integral
H
5z 2 −3z+2
C (z−1)3 dz,
(23)
where C is
any simple closed curve enclosing z = 1. Now by applying Cauchy
theorem, we find
I
C
2πi
5z 2 − 3z + 2
dz
=
(10)|z=1 = 10πi.
(z − 1)3
2!
(24)
An application of Cauchy integral formulae is the Cauchy inequal10
ity. Let f (z) =
P
anz n be analytic and bounded |f (z)| ≤ M on a
circle of radius r about the origin, then by Cauchy integral formulae
I
dz
= 2πi,
z
−
a
IC
dz
= 0, n ≥ 2
n
(z
−
a)
IC
(z − a)ndz = 0, n ≥ 1,
(25)
C
one arrives at
1
|an| = |
2π
Z
f (z)
1
dz|
≤
n+1
2π
|z|=r z
Z
|
|z|=r
f (z)
2πr
M
|dz
≤
M
=
.
z n+1
2πrn+1
rn
(26)
In summary, we have the Cauchy inequality |an|rn ≤ M .
An important application of Cauchy inequality is the Liouville’s
theorem : If f (z) is analytic and bounded in the entire complex
plane, then it is a constant. This can be understood easily from
|an| ≤
M
→ 0 when r → ∞
rn
(27)
for each n ≥ 1. Therefore f (z) = a0.
A famous application of Liouville’s theorem yields the fundamental theorem of algebra, which says that any polynomial P (z) =
Pn
v
v=0 av z with n > 0 and an 6= 0 has n roots : If P (z) has no
11
root, then 1/P (z) is analytic and bounded as |z| → ∞. Hence P (z)
is a constant by Liouville’s theorem! Therefore P (z) has at least
one root. Repeating this argument it is easy to see that P (z) has n
roots.
V.
SERIES AND SINGULARITIES
Most of the results of infinite series we introduced earlier apply to
the complex infinite series as well. For example, z = 0 is a removable
singularity of
1−cos z
z
z
because limx→0 1−cos
= 0. On the other hand,
z
ez does not have any singularity at finite z. However if one lets
z = 1/u, then e1/u has an essential singularity at u = 0. Therefore
ez has an essential singularity at z = ∞.
A.
Taylor Theorem
One of the most important theorem about infinite series is the
Taylor theorem
• If f (z) is analysis at all points inside and on a circle of radius R
with center at a, and if a + h is any point inside C, the Taylor
theorem says that :
h2 00
h3 000
f (a + h) = f (a) + hf (a) + f (a) + f (a) + ...... (28)
2!
3!
0
12
To prove Taylor theorem, one notices that by Cauchy theorem one
has
1
f (a + h) =
2πi
I
C
f (z)dz
.
z−a−h
(29)
Further we find
hn
1 h
h
h2
1
+ ... +
+
=
1+
+
z−a−h
z−a
z − a (z − a)2
(z − a)n
i
hn+1
+ ... .
(30)
(z − a)n+1
f (z)dz
back into f (a+h) =
C z−a−h
H
f (z)
n!
and use Cauchy theorems f n(a) = 2πi
C (z−a)n+1 dz, we immediately
Now put above expression of
1
z−a−h
1
2πi
H
have
h3 000
hn n
h2 00
f (a+h) = f (a)+hf (a)+ f (a)+ f (a)+....+ f (a)+Rn,
2!
3!
n!
(31)
0
where Rn is given by
hn+1
Rn =
2πi
I
C
f (z)dz
.
(z − a)n+1(z − a − h)
(32)
Notice one can easily show Rn → 0 as n → ∞. Hence we prove the
Taylor theorem.
13
Laurent Series
B.
If f (z) is analytic in an annular region r1 ≤ |z − a| ≤ r2, we can
generalize the Taylor series to a Laurent series. Let C1 and C2 be
concentric circles having center a and radii r1 and r2, respectively.
Also let a + h be any point in the annular region bounded by C1 and
C2. Then one can write f (a + h) as
1
f (a + h) =
2πi
I
C
f (z)dz
1
−
z − (a + h) 2πi
I
f (z)dz
,
z
−
(a
+
h)
C1
(33)
where C is any close curve in the angular region surrounding C1.
Then by expanding
1
z−a−h
in two different ways, one will reach
f (a + h) =
∞
X
anhn
(34)
−∞
with
1
an =
2πi
I
C
f (z)dz
.
(z − a)n+1
(35)
• If f (z) is analytic everywhere inside and on a simple closed curve
C except at z = a, which is a pole of order n. Then we have
a−n+1
a−1
a−n
+
+
...
+
+ a0 + a1(z − a)
(z − a)n (z − a)n−1
z−a
+ a2(z − a)2 + ....
(36)
f (z) =
14
• Further, one has
I
f (z)dz = 2πia1, .
(37)
C
Proof : By integration, we have
I
I
a−n
dz + ... +
f (z)dz =
n
(z
−
a)
C
C
= 2πia−1.
I
C
a−1
dz +
z−a
I
a0dz + ...
C
(38)
• and
i
1
dn−1 h
n
a1 = lim
(z − a) f (z) ,
z→a (n − 1!) dz n−1
(39)
here a1 is called the residue of the function at the pole z = a.
Proof : Multiplication by (z − a)n gives the Taylor series
(z −a)nf (z) = an +a−n+1(z −a)+...+a−1(z −a)n−1 +.... (40)
Taking the (n − 1)th derivative of both sides and letting z → a,
we arrive at
(n − 1)!a−1
i
dn−1 h
n
= lim n−1 (z − a) f (z) .
z→a dz
Let’s calculate the residues of the function f (z) =
poles z = 0, −2:
15
1
z(z+2)3
(41)
at the
1
1
• The residue at the simple z = 0 is limz→0 z · z(z+2)
3 = 8.
• The residue at the pole z = 2 of order 3 is
i
1
1
3
2) · z(z+2)3 = − 16
.
2
limz→−2 2!1 dzd 2
h
(z +
If f (z) is analysis within and on a simple closed curve C except
at a number of poles a, b, c, ... interior to C, then one has
I
h
i
f (z)dz = 2πi sum of residues of f (z) at poles a, b, c, .... .
C
(42)
Let’s apply above residue theorem to evaluate the integral
H
ez dz
C (z−1)(z+3)2 ,
where C is given by |z| = 10. First of all, the residue
h
i
ez
at z = 1 is given by limz→1 (z−1) (z−1)(z+3)2 = 16e . Next the residue
i
h
−3
ez
d
2
at z = −3 is given by limz→−3 dz (z + 3) (z−1)(z+3)2 = −5e
16 . Finally since both the poles z = 1, −3 are inside |z| = 10, one has
z
I
C
VI.
−3
e dz
5e
e
−
=
2πi
(z − 1)(z + 3)2
16
16
.
(43)
EVALUATION OF DEFINITE INTEGRALS
The materials we have covered so far regarding the complex anaylsis would be very useful for evaluating some definite integrals. To
demonstrate how these can be done, several theorems are required
• If |f (z)| ≤
M
Rk
for z = Reiθ , k > 1 and M is some constant.
16
then limR→∞
R
Γ f (z)dz
= 0, where Γ is the semicircular arc of
radius R in the upper half complex plane (counterclockwise).
Proof: one has
Z
Z
|
f (z)dz| ≤ |f (z)||dz|
Γ
Γ
Z π
M
πM
≤ kR
dθ = k−1 .
R
R
0
• If |f (z)| ≤
M
Rk
for z = Reiθ , k > 0 and M is some constant.
R
imz
f (z)dz
Γe
then limR→∞
(44)
= 0, where Γ is the semicircular arc
of radius R in the upper half complex plane (counterclockwiese)
and m > 0.
Proof: one has
Z
eimz f (z)dz| ≤
|
Γ
Z
Z0 π
|eimz f (z)||dz|
Γ
π
=
Z
Rdθ|eimR(cos θ+i sin θ)f (Reiθ )||d(Reiθ )|
M
Rk−1
0
Z π/2
M
=2
dθe−m sin θ k−1
R
Z0 π/2
M
≤2
dθe−m2Rθ/π k−1
R
0
πM
= k 1 − e−mR → 0 as R → ∞.
R
≤
dθe−m sin θ
17
(45)
Using above results, one can evaluate some complicated integrations which seem impossible. Let’s calculate the following integration
first
Z
0
Consider the integral
∞
1
dx.
x4 + 1
1
Γ z 4 +1 dz,
H
(46)
where Γ is the semicircular arc of
radius R in the upper hald complex plane (counterclockwise) and
the real axis from −R to R. Now by using above results, one see
that
I
dz
=2
4+1
z
Γ,R→∞
∞
Z
0
dx
.
x4 + 1
(47)
Now the z 4 + 1 = 0 when z = eiπ/4, ei3π/4, ei5π/4, ei7π/4. On the
other hand, only z = eiπ/4, ei3π/4 lies in the upper half complex
plane. Next, the residues of 1/(z 4 + 1) at z = eiπ/4, ei3π/4 are given
by e−3πi/4/4 and e−9πi/4/4, respectively. Hence
√
h
−3πi/4
−9πi/4 i
dz
e
e
π 2
=
2πi
+
=
.
4+1
z
4
4
2
Γ,R→∞
I
(48)
As a result, we find
Z
0
∞
√
π 2
1
dx =
.
x4 + 1
4
18
(49)
We want to calculate the following integral
∞
x2dx
.
2 + 1)2 (x2 + 2x + 2)
(x
−∞
Z
Consider the integral
H
C
z2
2
2
(z +1) (z 2 +2z+2)
(50)
dz, where C is the semicir-
cular arc Γ of radius R in the upper half complex plane (counterclockwise) and the real axis from −R to R. The poles of
z2
(z 2 +1)2 (z 2 +2z+2)
inside C are z = i of order 2 and z = −1 + i of order 1. Then the
residue at z = −1 + i is given by
z2
3 − 4i
lim (z + 1 − i) 2
=
.
z→−1+i
(z + 1)2(z 2 + 2z + 2)
25
(51)
Also, the residue at z = −i is found to be
d
lim
z→i dz
(z − i)2
z2
(z 2 + 1)2(z 2 + 2z + 2)
=
9i − 12
.
100
(52)
Hence we arrive at
I
C
z2
3 − 4i 9i − 12
7π
dz
=
2πi
+
=
.
(z 2 + 1)2(z 2 + 2z + 2)
25
100
50
(53)
19
On the other hand, we have
Z R
z2
x2
dz =
dx
2 + 1)2 (z 2 + 2z + 2)
2 + 1)2 (x2 + 2x + 2)
(z
(x
C
−R
Z
2
z
+
dz
2 + 1)2 (z 2 + 2z + 2)
(z
ZΓ ∞
x2
dx as R → ∞.
(54)
=
2 + 1)2 (x2 + 2x + 2)
(x
−∞
I
As a result, we reach
∞
x2dx
7π
=
.
2
2 2
50
−∞ (x + 1) (x + 2x + 2)
Z
Let’s see another example. Say, we want to calculate
(55)
R 2π
0
dθ
5+3 sin θ .
Let z = eiθ , then we have sin θ = (z − z −1)/(2i) and dz = ieiθ .
Hence
2π
Z
0
dθ
=
5 + 3 sin θ
I
C
2dz
,
3z 2 + 10iz − 3
(56)
here C is the circle of unit radius with center at the origin (moving
in counterclockwise). The poles of
1
3z 2 +10iz−3
are simple poles at
−3i, −i/3 and only −i/3 is within C. Further, the residue at z =
−i/3 is given by 1/(4i). Therefore we have
Z
0
2π
dθ
1
π
= 2πi = .
5 + 3 sin θ
4i 2
20
(57)
Our final example is to determine the integral
H
0. To calculate the integral, one can consider C
R ∞ cos mx
0
x2 +1
eimz dz
,
z 2 +1
dx,
m>
here C is the
same contour as in our first example. Carrying out similar conisderation will lead to
Z
R
cos mxdx
+i
2+1
x
−R
Z
R
sin mxdx
+
2+1
x
−R
eimz dz
= πe−m.
2
Γ z +1
Z
(58)
Hence by taking R → ∞, one has
Z
0
∞
cos mx
π −m
dx
=
e .
x2 + 1
2
(59)
You might want to show
Z
∞
π
sin x
dx = .
x
2
Z0 ∞ p−1
x
π
dx =
, 0 < p < 1.
1+x
sin px
0
Z 2π
cos 3θ
π
dθ = .
(60)
5
−
4
cos
θ
12
0
R ∞ sin x
Let’s calculate 0 x dx. Let C be the contour consisting of
semicircles of radii r and R in the upper half x-y plane (r < R and
the orientation of semicircle with radius r (R) is moved clockwise
(counterclockwise)), and the part of x-axis from −R to −r and
from r to R. Since the singularity of eiz /z, which is at z = 0, is not
21
within the region bounded by C, one has
eiz
dz = 0,
z
I
C
(61)
or one has
Z
−r
−R
eix
dx +
x
Z
Γr
eiz
dz +
z
R
Z
r
eix
dx +
x
eiz
dz = 0.
ΓR z
I
(62)
R −r ix
Further, by replacing x by −x in the integral −R ex dx, one has
R r e−ix
R R e−ix
R −r eix
dx
=
dx
=
−
R x
r
−R x
x dx. Consequently, one reaches
Z
Γr
eiz
dz +
z
R
Z
r
eix
dx −
x
R
Z
r
e−ix
dx +
x
I
eiz
dz −
z
Z
eiz
dz = 0,
z
ΓR
(63)
eiz
dz,
ΓR z
(64)
or
R
Z
2i
r
sin x
dx =
x
Z
Γr1
here Γr1 is the counterclockwise semicircle with radius r (and cenR eiz
tered at the origin). The integral ΓR z dz is zero. Further, by going
into polar coordinate, one can show
Z
Γr1
eiz
dz = lim
r→0
z
Z
π
iθ
ieire dθ = iπ.
(65)
0
As a result, one arrives at
Z
0
∞
π
sin x
dx = .
x
2
22
(66)
A.
Laplace transform
Let
R∞
0
f (t)e−stdt = f (s), where f (s) is a given rational function
with numerator of degree less than that of the denominator. If C
is a simple closed curve enclosing all the poles of f (s), we can show
that
1
F (t) =
2πi
I
eztf (z)dz.
(67)
C
f (s) is called that Laplace transform of F (t) and F (t) is the inverse
Laplace transform of f (s). Laplace transform is very useful in solving
ordinary differential equations.
Example: Let f (s) =
s
s2 +1
. Then since the poles of
z
z 2 +1
are at
z = ±i, one has
i
1 it
−it −i
−it
F (t) = e
+e
=
e +e
= cos t.
2i
−2i 2
it
VII.
(68)
BRANCH LINE
Let w =
√
z. Let a point A moves along the circle |z| = 1. When
A has returned to its starting point, its image point has not yet
returned. When A has returned for the second time, its image point
has returned for the first time. In other word, w is not a single-
23
valued function of z, but is a double-valued function of z. To ensure
a single-valued function, we must restrict θ, say choose 0 ≤ θ < 2π.
√
This will represent one branch of the double-valued function z = z.
In continuing beyond this interval, one is on the second branch. The
point z = 0 about which the rotation is taking place is called the
√
branch point. Equivalently, we can ensure that f (z) = z will be
single-valued by agreeing not to cross the line x ≥ 0 which is called
a branch line.
VIII.
CONFORMAL MAP
Let A = (2, 1), B = (4, 1) and C = (4, 3) be 3 points in z-plane.
Then under the transformation w = z 2, we have A0 = (3, 4), B 0 =
(15, 8) and C 0 = (7, 24) in the w-plane. The lines conecting A, B
and AC are given by y = 1 and y = x − 1, respectively. Notice
the angle between AB and AC is π/4. As a result of the map
w = z 2, the curves A0B 0 and A0C 0 in the w-plane are determined by
v 2 = 4(1+u) and u2 = 2v +1, respectively. The slope of the tangent
to the curves A0B 0 and A0C 0 at (3, 4) are given by
dv
du |(3,4)
dv
du |(3,4)
= 1/2 and
= 3, respectively. Hence the angle between the two curves at
24
A0 is given by
tan θ =
m2 − m1
= 1,
1 + m1m2
θ=
π
.
4
(69)
One can also show that the angle between CA(BA) and CB(BC)
are the same as the angle between C 0A0(B 0A0) and C 0B 0(B 0C 0).
Such map that preserves angles locally is called conformal map. An
analytic function f (z) which satisfies f 0(z0) 6= 0 at z0 is a conformal
map at z0.
25