Complex Analysis Lecturenotes Nai-Sher Yeh

Complex Analysis Lecturenotes Nai-Sher Yeh

December 22; 2010

Complex Analysis 2 December 22; 2010

COMPLEX ANALYSIS

Quizes : 0%

Midterm : 50%

Final : 50%

Instructor : Nai-Sher Yeh, assistant professor Office : MA315

Phone : Ext. 2460

Complex Analysis 3

1 Complex numbers

1.1 De nitions and symbols

Def. A complex number z is de ned as z = a + bi; where a; b 2 R; i =p

1;

and a Re z : real part of z;

b Im z : imaginary part of z:

The collection of all complex numbers is denoted by C^. 1.1.1 Symbols of complex numbers

8z 2 C;

z = a + bi;

z a bi : the ”complex conjugate” of z;

jzj p

a² + b² : the ”modulus”(absolute value) of z:

1.1.2 Addition and multiplication rules

8z¹; z₂ 2 C; z¹ = a + bi; z₂ = c + di; then we have:

1): z₁ + z₂ = (a + c) + (b + d)i;

2): z = Re z i Im z;

3): z₁ z₂ = (ac bd) + (ad + bc)i;

4): ^z_z¹

2 = (a+bi)(c di)

(c+di)(c di) = ^ac+bd_c2+d2 + ^{bc ad}_c2+d2i:

Commutative rule for addition and multiplication is also valid.

Complex Analysis 4 Use the above notations and consider the following:

Ex. 1

jzj² = a² + b² = z z:

Ex. 2

z₁ + z₂ = (a + c) (b + d)i = z₁ + z₂: Ex. 3

z₁z₂ = (ac bd) (ad + bc)i = z₁ z₂: Ex. 4

z₁

z₂ = ac + bd c² + d²

bc ad

c² + d² i = z₁ z₂:

Complex Analysis 5 1.2 Polar coordinate and Cartesian coordinate

The complex numbers can be viewed as vectors. This is to consider the real part as x-axis and the imaginary part as y-axis.

And we call such a 2-dimensional plane the ”complex plane.”

Figure.

It is also natural to think of a point in terms of polar coordinate. Therefore, we see that

r = jzj p

a² + b² ( 0) ; a = r cos = Re z;

b = r sin = Im z;

z = r (cos + i sin ) = reⁱ :

The last notation for z can be veri ed by using simple geometry and Maclaurin series for eⁱ ; cos and sin :

Def. 8z 2 C; z = reⁱ : Then the ”argument of z;” denoted by arg z; means the radial value of angle of z; i. e.

arg z :

1.3 Basic properties of complex numbers

By viewing complex numbers as vectors, we may have the

Complex Analysis 6 following properties:

Figure.

1).

jz¹ + z₂j jz¹j + jz²j : triangular inequality.

2).

jz¹ + z₂j² jz¹j² + 2 jz¹z₂j + jz²j² :

pf. jz¹ + z₂j² = (z₁ + z₂) (z₁ + z₂) = (z₁ + z₂) (z₁ + z₂)

= z₁z₁ + z₁z₂ + z₂z₁ + z₂z₂:

* z¹z₂ + z₂z₁ = z₁z₂ + z₁z₂ = 2 Re z₁z₂ 2 jz¹z₂j ; ) z¹z₁ + z₁z₂ + z₂z₁ + z₂z₂ z₁z₁ + 2 jz¹z₂j + z²z₂:

3).

jz¹z₂j = jz¹j jz²j ; ^and jzj = jzj : 4).

jz¹ + z₂j² (jz¹j + jz²j)² : Schwartz inequality.

pf. The result is immediate followed by the combination of 2).

and 3).

5). Theorem:

Complex Analysis 7 8z; ! 2 C; z = re^{i and} ! = e^i': Then we have

z = r (cos + i sin ) ; 8 2 R (De Moivre), z! = r (cos ( + ') + i sin ( + ')) ;

and z

! = r

(cos ( ') + i sin ( ')) ; if 6= 0:

pf. The result is obvious followed by the polar-exponential notations of z and !: Note that, from the above results, we also have

arg z! = arg z + arg !;

and arg z

! = arg z arg !:

Ex. In case z = reⁱ ; n 2 N; ^{we nd that} zⁿ = rⁿ(cos n + i sin n ) : Ex. In case ! = e^i'; = _n¹; n 2 N: ^Then

! = pⁿ

! = pⁿ

cos ' + 2k

n + i sin ' + 2k

n ; 8k 2 Z:

Complex Analysis 8 1.4 Inversion

Def. Given ; 2 C; we say that and are ”symmetric to the circle jzj = R^{” if(f.)} j j = R:

Def. Given ; 2 C; we say that is the ”inversion of (6= 0)^{” if(f.)} j j = 1:

Geometric explanation:

Figures.

For 6= 0; to construct a point = 1= ; consider the following:

i). j j = j1= j = j1= j ; ^and

ii). arg = arg 1 arg = 0 + arg = arg (Since arg = arg :)

Then we may see from the above gures, that iii). j j < 1; 4O P v 4OP ;

iv). j j > 1; 4O P v 4OP ; ^{and that} v). j j = 1; = :

Complex Analysis 9

2 Limits in complex plane

2.1 Principle of nested regions

Theorem 1 (Principle of nested intervals)

Given fI^kg¹_k=1 : sequence of closed intervals on R; ^and i): I₁ I₂ ::: I_n I_n+1 :::

ii): I_n = [a_n; b_n] ; lim

n!1(b_n a_n) = 0:

)

Then ¹\

n=1I_n contains exactly 1 point.

pf. Figure.

It is obvious that {a_k}: an increasing bounded (a_k b₁; 8k⁾ sequence; therefore,

sup

k

a_k = lim

k!1 a_k = a exists by monotonic sequence theorem.

We also see that {b_k}: an decreasing bounded (b_k a₁; 8k⁾ sequence; therefore, similarly to the above, we have

infk b_k = lim

k!1b_k = b:

Assume that a 6= b; ^then 9c 2 (a; b) ^{such that} c 2 I^k; 8k: ^Hence a 6= sup

k

a_k and b 6= inf

k b_k; which is a contradiction. ]

Complex Analysis 10 Theorem 2 (Principle of nested rectangles)

Given fr^kg¹_k=1 : sequence of closed intervals on R^{2 (or} C⁾; and

i): r₁ r₂ ::: r_n r_n+1 :::

ii): l_k : the diagonal length of r_k; lim

k!1l_k = 0:

)

Then ¹\

k=1r_k contains exactly 1 point.

pf. Figure. r_k = I_k J_k

Set I_k : the projection of r_k onto x axis, and J_k : the projection of r_k onto y axis. Then fI^kg; fJ^kg are sequences of nested intervals on x axis and y axis, respectively. Note that the length of either I_k or J_k must l_k:

Now, since 9!a; b ^{such that 1}\

k=1I_k = fag ^{and that 1}\

k=1J_k = fbg by the above theorem, i. e.

1\

k=1r_k = f(a; b)g : ]

Remark 1 In fact, we may use the above theorems and induction to show that is also true on any nite dimensional space R^{n, that} the union of n dimensional nested rectangles contains also ex-

Complex Analysis 11 actly one point.

Complex Analysis 12 2.2 Limit points

Def. We say that the number is a ”limit point” of a sequence of complex numbers fzⁿg¹n=1 if(f.) 8" > 0; the deleted

neighborhood

D_" ( ) = fzj 0 < j zj < "g of contains points in fzⁿg¹n=1 :

Ex. 1, 3, 5, 7, 9, ... has no limit point.

Ex. 1, ₂ⁱ; ¹₃; ²₃i; ¹₅; ⁴₅i; ...has limit points 0 and i

Def. A complex sequence fzⁿg is bounded if(f.) 9M > 0 such that jzⁿj M; 8n^.

Theorem 3 (Bolzano-Weierstrass)

Every bounded complex sequence fzⁿg has a limit point.

pf. Figure.

Take a rectangle r₁ ^{such that} z_n 2 r¹; 8n: Next, divide r₁ ^into 4 equal parts—one of them must contain in nitely many terms of fzⁿg :

Continue doing it in this way, we will have a sequence of

Complex Analysis 13 rectanglesfr^kg ; ^where

r₁ r₂ ::: r_k :::::

,and each r_k contains in nitely many points of fzⁿg : By principle of nested rectangles, there must be exactly a point such that

2 ¹\

k=1r_k: Then is the limit point.

Complex Analysis 14 2.3 Convergence of complex sequences

Def. fzⁿg is convergent to the limit if(f.)

8" > 0; 9N > 0 : 8n > N ) jzⁿ j < ":

This also means that

nlim!1 z_n = , or z_n ! ^as n ! 1:

Theorem 4 (Cauchy criterion)

nlim!1 z_n = iff.

8" > 0; 9N > 0 : 8m; n > N ) jzⁿ z_mj < ":

pf. ()) Obviously. (Homework)

(() First of all, let's claim that there indeed is a limit point:

Choose " = 1; 9N⁰ > 0 such that 8 n > N⁰ ) jzⁿ z_N0+1j < 1: By triangle inequality, we nd that

jzⁿj jzⁿ z_N0+1j + jz^N0+1j < 1 + jz^N0+1j :

So jzⁿj M = maxf1 + jz^N0+1j ; jz¹j ; jz²j ; ^..., jz^N0jg ; ^{i. e.}

fzⁿg is bounded. By Bolzano-Weierstrass theorem, we nd that there is a limit point for fzⁿg :

Secondly, by hypothesis, we have

8" > 0; 9N > 0 : 8m; n > N ) jzⁿ z_mj < ";

Complex Analysis 15 and from the fact that fzⁿg has a limit point, say ; we obtain that

9m⁰ > N ) j z_m0j < ":

Therefore, we conclude that

8" > 0; 9N > 0 : 8n > N

) jzⁿ j < jzⁿ z_m0j + jz^m0 j < " + " = 2";

i. e.

nlim!1 z_n = :

Complex Analysis 16 2.4 Riemann sphere and extended complex plane

Figure.

Consider the mapping M : ! P

; where p 2 : the complex plane and p⁰ 2 P

: the sphere. We may show that the mapping M : ! P

n fN g is a bijection, where N ^{is the}

”north pole” of P :

Def. The bijection M : ! P

n fN g is called the

”stereographic projection,” and the sphere P

is called ”Riemann sphere.”

Remark 2 M : ! P

n fN g is a ”homeomorphism” in topol- ogy.

Def. The ”ideal point” of the complex plane is ”in nity”(1^.) Def. The ”extended complex plane” is the complex plane together with ideal point ( i. e. [ f1g :⁾

Remark 3 The mapping M : [f1g ! P becomes a bijection when M (1) = N ^{is de ned.}

Complex Analysis 17 Def. lim

n!1 z_n = 1 ^if

8M > 0; 9N > 0; ^{such that} 8n > N ) jzⁿj > M:

Def. E is said to be the ”exterior of a circle” if(f.) E is a neighborhood of 1 ^{in case} E is regarded as a subject on the extended complex plane. We may also say that E is a deleted neighborhood of 1 ^{in case} E is regarded as a subset on the complex plane.

Complex Analysis 18

3 Complex functions

3.1 Basic concept

Def. The complex valued function f : E ! eE; E; eE C ^{is said} to be a ”single-valued function” if it assigns each value z 2 E to a unique complex number ! on E: fe is otherwise called a

”multi-valued function.”

Remark 4 We will see later that both of these de nitions are well- de ned if we consider the range being a multi-branch Riemann surface instead of the usual one-layer complex plane.

**Note: we usually separate a complex valued function into the real part function and imaginary part function, i. e. for z = x + yi; we write

f (z) = ! = u + vi;

where u = u (x; y) and v = v (x; y) : They are both real-valued functions.

Ex. Split f (z) = z² into u (x; y) + iv (x; y) : Sol. z² = x² y² + 2xyi: Therefore we nd

u (x; y) = x² y² and v (x; y) = 2xy:

3.2 Curves and domains

Def. Given x = x (t) and y = y (t) : real-valued functions and

Complex Analysis 19 x (t) ; y (t) 2 C [a; b] : Then the function

z (t) = x (t) + iy (t) ; 8t 2 [a; b]

is called a curve on complex plane.

Figure.

Def. z (a) is called the ”initial point,” and z (b) is called the

”terminal point,” according to the above de nition. We say that the curve is ”closed” if z (a) = z (b) :

Def. A curve is said to be ”simple” if z (t) is 1-1.

Figures.

Remark 5 Geometrically speaking,a simple curve is a curve with- out knot.

Def. A simple closed curve is called a ”Jordan curve.”

Def. For a Jordan curve, we de ne that the counterclockwise direction the ”positive direction.”

Complex Analysis 20 Figure.

Def. Consider some E C: ^Then E is said to be ”(arcwise or pathwise) connected” if each pair of points z₁ and z₂ on E can be joined by a curve lying entirely on E:

Def. Consider some G C: ^Then G is a ” domain” if G is both open and connected.

Def. G is called the ” closure of G;” which is a domain G together with its boundary points. We also call G a ”closed domain.”

Def. Ge is a called a ” region” if Ge is a domain G together with part of its boundary points. (G : Ge wigle.)

Def. A range G is said to be ”simply connected” if for any given Jordan curve C ⁱⁿ G; the interior of C is also connected in G:

Complex Analysis 21 3.3 Limit and continuity

Def. A complex function f is de ned on a domain G; and z₀ 2 G: ^Then

zlim!z⁰ f (z) = ( 2 C^{) or} f (z) ! ^as z ! z⁰ means

8" > 0; 9 > 0 : 0 < jz z₀j < ) jf (z) j < ":

Furthermore, we say that f is ”continuous at z₀” if f (z₀) = : Def. A complex function f is said to be ”continuous on G” if it is continuous at every point on G:

Theorem 5 A complex functionf (z) = u (x; y)+iv (x; y) ; where z₀ = x₀ + iy₀ and !₀ = u₀ + iv₀: Then we nd that

zlim!z⁰ f (z) = !₀ iff.

lim

(x;y)!(x⁰;y₀)u (x; y) = u₀ and lim

(x;y)!(x⁰;y₀)v (x; y) = v₀: pf. ()⁾

* lim_z

!z⁰ f (z) = !₀

) 8" > 0; 9 > 0 : 0 < jz z⁰j < ) jf (z) !₀j < ":

Complex Analysis 22 i. e.

0 <

q

(x x₀)² + (y y₀)² <

) ju (x; y) u₀j jf (z) !₀j < " ^and jv (x; y) v₀j jf (z) !₀j < ":

((^{) We have}

8" > 0; 9 > 0 : 0 <

q

(x x₀)² + (y y₀)² <

) ju (x; y) u₀j < " ^and jv (x; y) v₀j < ":

Therefore, it is clear that jz z₀j =

q

(x x₀)² + (y y₀)²; and that for 0 < jz z₀j <

) jf (z) !₀j ju (x; y) u₀j + jv (x; y) v₀j < 2":

i. e. lim

z!z⁰ f (z) = !₀:

Properties of limits:

Consider f; g : complex functions both de ned on a domain G: There is some z₀ 2 G ^and

zlim!z⁰ f (z) = as well as lim

z!z⁰ g (z) = : Then we have:

1). lim

z!z⁰ (f (z) +g (z)) = + ;

Complex Analysis 23 2). lim

z!z⁰ (f (z) =g (z)) = = provided that 6= 0:

3). If f is continuous at z₀ and g is continuous at f (z₀) ; then g f is continuous at z₀.

Def. If f is de ned on a region Ge and for some z₀ 2 fG:

Then f is ”continuous at z₀” means that 8" > 0; 9 > 0 ^{such that} jz z₀j < ^, z 2 eG ) jf (z) f (z₀)j < ": The notation of this is

zlim!z⁰ z2 eG

f (z) = f (z₀) :

Def. If a complex function f is ”continuous at z₀ of a curve C”, that means

zlim!z⁰ z2C

f (z) = f (z₀) :

Def. A complex function f is ”continuous on C^{” if} f is continuous at every point of C ^.

Complex Analysis 24 3.4 Uniform continuity

Def. f is de ned on a set E C: ^Then f is ”uniformly continuous” on E if

8" > 0; 9 > 0 : jz z⁰j < ; 8z; z⁰ 2 E ) jf (z) f (z⁰)j < ":

Ex. Set f (z) = _{1 z}¹ and the domain of f is de ned as G = fzj jzj < 1g : ^{Show that} f is not uniformly continuous on G:

pf. Choose "₀ = 1; and set z = 1 _n¹; z⁰ = 1 _n+1¹ : Then we nd that jz z⁰j = _n(n+1)¹ < ; but jf (z) f (z⁰)j = jn (n + 1)j = 1 = "⁰: Therefore it is not uniformly continuous.

Remark 6 When you want to show certain function being ”not uniformly continuous,” what you have to do is to consider the negation of the original de nition, i. e.

9" > 0 ^{s. t.} 8 > 0; jz z⁰j < ; 9z; z⁰ 2 E but jf (z) f (z⁰)j ":

Theorem 6 (Heine-Borel)

G : a bounded closed domain, and consider 8z 2 G ^{is the} center of a disk D_z: Then we nd that G can be covered by a nite number of D_z's.

Complex Analysis 25 pf. Simply consider a rectangle r₁ G; and assume that G cannot be covered by nitely many disks D_z: Then we divide r₁ into 4 rectangles, and at least one of these 4 rectangles contains part of G that cannot be covered by nitely many D_z: Let's set this rectangle (one of them) as r₂:

Continue doing it in this way, we obtain a set of rectangles frⁿg¹n=1, r₁ r₂ ::: r_n :::; and r_n \ G cannot be covered by nitely many disks D_z; 8n; ^{i. e.} r_n contains in nitely many points of G; 8n:

From the nested rectangle theorem, we nd that 9!z⁰ 2

1\

n=1r_n: Then it is clear that this point z₀ must be a limit point of G—since

8" > 0; 9N > 0 : r^N fz j jz z₀j < "g ;

then obviously the arbitrary neighborhood (of z₀) contains

in nitely many points of G: Therefore, because of the closedness of G; we also nd z₀ 2 G: Let's set this neighborhood

fz j jz z₀j < "g D_z0 (") :

Finally, we got a contradiction—for we may nd that there is a certain suf ciently large N such that r_N D_z0 (") ; and r_N shouldn't be able be covered by nitely many disks D_z yet it is covered by a single disk D_z0 (") :

Theorem 7 f (z) is continuous on a bounded closed domainG ) f (z) is uniformly continuous on G:

Complex Analysis 26 pf. Homework!

Complex Analysis 27

4 Differentiation in the complex plane

4.1 The derivative of a complex function

Def. f (z) : de ned on a domain G; z 2 G: Then we have the following de nitions:

i).

f⁰(z) = lim

4z!0

f (z + 4z) f (z) : the derivative of f at z: 4z

ii). f is ”analytic” on a domain G if f is differentiable at each point of G; i. e. f⁰ (z) exists on G:

iii). f is analytic at z₀ if f is differentiable at each point of some neighborhood of z₀:

Ex. f (z) = z³ : analytic on the complex plane, and f⁰ (z) = 3z²:

Ex. f (z) = Re z : continuous on the whole z-plane but nowhere differentiable. Reason:

4z!0lim

f (z + 4z) f (z)

4z = lim

4z!0

Re (z + 4z) Re (z) 4z

= lim

4z!0

Re 4z 4z

: doesn't exist (you may choose different paths to attain 4z ! 0 and nd different values.)

Ex. f (z) = z: Show that f is not differentiable.

Complex Analysis 28 pf. Figure.

Note that

4z!0lim

f (z + 4z) f (z)

4z = lim

4z!0

z + 4z z

4z = lim

4z!0

4z 4z

= 1; if 4z = 4x;

1; if 4z = i4y:

Therefore f⁰(z) doesn't exist.

Properties: Given differentiable complex functions f, g and constants ; 2 C:

1). ( f + g)⁰ = f⁰ + g⁰: (linearity) 2). (f g)⁰ = f⁰g + f g⁰:

3). (f =g)⁰ = (f⁰g f g⁰) =g²; provided that g 6= 0:

4). Df (g (z)) _dz^d f (g (z)) = ^df_dg ^dg_dz = f⁰ (g (z)) g⁰ (z) : (chain rule)

5). f (z) = z ; 2 R ) f⁰ (z) = z ¹: (power rule) Ex. A complex polynomial function P (z) = Pn

k=0c_kz^k; c_k 2 C; 8k is analytic over the whole complex plane (by linearity and power rule.)

Def. A ”complex increment” of a complex function

Complex Analysis 29

! = f (z) differentiable at z is de ned as

4! = f (z + 4z) f (z) = [f⁰ (z) + "] 4z;

where lim_4z!0 " = 0:

Def. A ”complex differential” of a complex function

! = f (z) differentiable at z is de ned as df f⁰ (z) dz:

Complex Analysis 30 4.2 The Cauchy-Riemann equations

Def. The ”Cauchy-Riemann equations” of a complex function

! = f (z) differentiable at z = x + iy is de ned as

@u

@x = @v

@y ^and

@u

@y = @v

@x

at (x; y) ; where f = u + iv, u = u (x; y) and v = v (x; y) : Note: Quite a few ideas on complex plane come from the 2-dimensional space on R²: Now, for a complex function f; there are the real part function u and imaginary part function v; and both of them the 2-variable functions. When consider a real valued function u (x; y) being differentiable at (x; y) ; we mean that

u (x + 4x; y + 4y) u (x; y)

= (u_x + "₁) 4x + (u^y + "₂) 4y : the increment of u and "₁; "₂ satisfy that

lim

(4x;4y)!(0;0)"_j = 0; j = 1; 2:

This can also be applied to v (x; y) :

Theorem 8 ! = f (z) = u + iv is a differentiable function at the point z = x + iy iff. the functions u and v are differentiable at the point (x; y) and satisfy the Cauchy-Riemann equations.

pf. ()⁾ * f is differentiable at z; 4! = (f⁰ (z) + ") 4z;

Complex Analysis 31 where lim_4z!0 " = 0: Now,

4z = 4x + i4y;

4! = 4u + i4v; ^and

" = "₁ + i"₂;

where lim₍4x;4y)!(0;0) "_j = 0; j = 1; 2: Let f⁰(z) = + i ; then we have

4u + i4v = [( + i ) + ("¹ + i"₂)] (4x + i4y) ; therefore

4u = 4x 4y + "¹4x "₂4y;

4v = 4x + 4y + "²4x + "¹4y;

i. e. u and v are differentiable at (x; y); and = u_x; = u_y; also = v_y; = v_x: This implies that

@u

@x = @v

@y ^and

@u

@y = @v

@x: ((^{) We have}

4u = u (x + 4x; y + 4y) u (x; y)

= (u_x + "₁) 4x + (u^y + "₂) 4y; ⁽¹⁾ where

lim

(4x;4y)!(0;0)"_j = 0; j = 1; 2 and

Complex Analysis 32 4v = v (x + 4x; y + 4y) v (x; y)

= (v_x + ₁) 4x + (v^y + ₂) 4y; ⁽²⁾ where

lim

(4x;4y)!(0;0) k = 0; k = 1; 2:

From Cuachy-Riemann equations and equation (1), we obtain 4u = (u^x + "₁) 4x + ( v^x + "₂) 4y;

similarly, we have

4v = (v^x + ₁) 4x + (u^x + ₂) 4y:

Then the complex increment 4! has the form

4! = 4u + i4v = (u^x + iv_x) (4x + i4y) + ("₁ + i ₁) 4x + ("² + i ₂) 4y;

and furthermore, 4!

4z = (u_x + iv_x) + ("₁ + i ₁) 4x

4z + ("₂ + i ₂) 4y 4z: Note that

j"¹ + i ₁j 4x

4z j"¹j + j 1j ! 0 ^as (4x; 4y) ! (0; 0) ; ^and j"² + i ₂j 4y

4z j"²j + j 2j ! 0 ^as (4x; 4y) ! (0; 0) :

Complex Analysis 33 i. e.

4z!0lim

4!

4z f⁰ (z) = u_x + iv_x(= v_y iu_y):

Ex. f (z) = z²: Use the above theorem to verify if f is differentiable on the complex plane and nd its derivative if it is differentiable.

Sol. Let z = x + iy; and then f (z) = x² y² + i (2xy) : In this case,

u (x; y) = x² y² and v (x; y) = 2xy:

Then we see that

u_x = 2x = v_y and u_y = 2y = v_x:

Also, u and v are continuous and differentiable everywhere on R²: Hence according to the above theorem, f (z) is differentiable at any point on C:

Complex Analysis 34 4.3 Conformal mapping

Discussion: Given a curve C : z = z (t) (= x (t) + iy (t));

t 2 [a; b] : The tangent vector of C ^at z (t) is z⁰ (t) = x⁰(t) + iy⁰ (t) ;

and note that the direction of a tangent line of the curve C ^is determined by arg z⁰ (t) :

Figure. 1

(One curve to one curve.) Figure. 2

(Two intersecting curves to two intersecting curves.) Consider a curve : !(t) = f (z(t)); t 2 [a; b] : ^When given a point t₀ on [a; b] ; such that z⁰(t₀) 6= 0 ^and f⁰ (z (t₀))(=

Complex Analysis 35 f⁰ (z₀))6= 0; ^{we nd that}

!⁰ (t₀) = f⁰ (z₀) z⁰ (t₀) 6= 0;

arg !⁰ (t₀) = arg f⁰ (z₀) + arg z⁰ (t₀) ; also

' = arg f⁰ (z₀) + and ' = arg f⁰ (z₀) + :

Therefore, it is clear that ' ' = ; which means the mapping f ”preserves” the angle between tangent lines.

Def. A mapping f : C ! C is said to be ”conformal” at all points z with f⁰(z) 6= 0:

Def. A mapping by a continuous function which preserves only magnitude of angles between any two curves passing through a point z₀ is said to be ”isogonal at z₀.”

Ex. ! = f (z) = z² : not conformal at z = 0:

Figure.

Note that f⁰(z) = 2z; f is conformal at all points z such that z 6= 0: Also it is clear that arg ! = 2 arg z: When we consider the line z = t (1 + i) ; t 2 R; we see that f (z) = f (t (1 + i)) = 2it²; which is not conformal at 0; as shown in the above gure.

Complex Analysis 36 Ex. ! = z : isogonal but not conformal at every point of z:

Figure.

Note that the orientation of the angles are opposite to each other after mapping.

Complex Analysis 37

5 Integration in the complex plane

5.1 Line integral in 2-dimensional spaces

Def. A curve C on a 2-dimensional space is de ned as C : P (t) = (x (t) ; y (t)) ; a t b;

where x (t) and y (t) are continuous functions on [a; b] :

Def. A curve C is said to be ”smooth” if P⁰(t) 6= 0 ^and x (t) ; y (t) 2 C¹[a; b] :

Def. Given a real-valued function f (x; y) de ned on a smooth curve C. Then the ”line integral of f (x; y) over smooth curve C with respect to x” is de ned as follows:

Consider = ft^kj a = t⁰ < t₁ < ::: < t_n = bg : ^partition of [a; b] ;

c_k 2 [t^{k 1}; t_k] ; k = 1; 2; :::n;

x_k = x(t_k) x (t_{k 1}) ;

P_k = P (c_k) = (x (c_k) ; y (c_k)) 2 C:

ZThen

C

f (P ) dx = Z

C

f (x; y) dx lim

k k!0

Xn k=1

f (P_k) x_k;

Complex Analysis 38 or

8" > 0; 9 > 0; ^{such that} 0 < k tk <

)

Xn k=1

f (P_k) x_k

Z

C

f (P ) dx < ":

Theorem 9 f is continuous on a smooth curve C^{. Then} Z

C

f (P ) dx exists and equals to

Z b a

f (x (t) ; y (t)) x⁰(t) dt:

pf. The smooth curve C : P (t) = (x (t) ; y (t)) ; x(t);

y(t) 2 C¹[a; b] : Then consider the partition of [a; b]:

= ft^kj a = t⁰ < t₁ < ::: < t_n = bg ; c_k 2 [t^{k 1}; t_k] ; k = 1; 2; :::n; and

=

Xn k=1

f (x (c_k) ; y (c_k)) (x (t_k) x (t_{k 1})) ;

which is the Riemann sum of the line integral. Again, look at the Riemann sum S of

Z _b

a

f (x (t) ; y (t)) x⁰ (t) dt : S =

Xn k=1

f (x (c_k) ; y (c_k)) x⁰ (c_k) t_k: