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Complex Analysis

George Cain

Chapter One - Complex Numbers
1.1 Introduction
1.2 Geometry
1.3 Polar coordinates

Chapter Two - Complex Functions
2.1 Functions of a real variable
2.2 Functions of a complex variable
2.3 Derivatives
Chapter Three - Elementary Functions
3.1 Introduction
3.2 The exponential function
3.3 Trigonometric functions
3.4 Logarithms and complex exponents
Chapter Four - Integration
4.1 Introduction
4.2 Evaluating integrals
4.3 Antiderivatives

Chapter Five - Cauchy's Theorem
5.1 Homotopy
5.2 Cauchy's Theorem
Chapter Six - More Integration
6.1 Cauchy's Integral Formula
6.2 Functions defined by integrals
6.3 Liouville's Theorem
6.4 Maximum moduli
Chapter Seven - Harmonic Functions
7.1 The Laplace equation
7.2 Harmonic functions
7.3 Poisson's integral formula

Chapter Eight - Series
8.1 Sequences
8.2 Series
8.3 Power series
8.4 Integration of power series
8.5 Differentiation of power series
Chapter Nine - Taylor and Laurent Series
9.1 Taylor series
9.2 Laurent series

Chapter Ten - Poles, Residues, and All That
10.1 Residues
10.2 Poles and other singularities
Chapter Eleven - Argument Principle
11.1 Argument principle
11.2 Rouche's Theorem

----------------------------------------------------------------------------
George Cain
School of Mathematics
Georgia Institute of Technology
Atlanta, Georgia 0332-0160
cain@math.gatech.edu
Chapter One

Complex Numbers
1.1 Introduction. Let us hark back to the first grade when the only numbers you knew
were the ordinary everyday integers. You had no trouble solving problems in which you
were, for instance, asked to find a number x such that 3x пЂЅ 6. You were quick to answer
вЂќ2вЂќ. Then, in the second grade, Miss Holt asked you to find a number x such that 3x пЂЅ 8.
You were stumpedвЂ”there was no such вЂќnumberвЂќ! You perhaps explained to Miss Holt that
3пѓќ2пѓћ пЂЅ 6 and 3пѓќ3пѓћ пЂЅ 9, and since 8 is between 6 and 9, you would somehow need a number
between 2 and 3, but there isnвЂ™t any such number. Thus were you introduced to вЂќfractions.вЂќ

These fractions, or rational numbers, were defined by Miss Holt to be ordered pairs of
integersвЂ”thus, for instance, пѓќ8, 3пѓћ is a rational number. Two rational numbers пѓќn, mпѓћ and
пѓќp, qпѓћ were defined to be equal whenever nq пЂЅ pm. (More precisely, in other words, a
rational number is an equivalence class of ordered pairs, etc.) Recall that the arithmetic of
these pairs was then introduced: the sum of пѓќn, mпѓћ and пѓќp, qпѓћ was defined by

пѓќn, mпѓћ пЂ« пѓќp, qпѓћ пЂЅ пѓќnq пЂ« pm, mqпѓћ,

and the product by

пѓќn, mпѓћпѓќp, qпѓћ пЂЅ пѓќnp, mqпѓћ.

Subtraction and division were defined, as usual, simply as the inverses of the two
operations.

In the second grade, you probably felt at first like you had thrown away the familiar
integers and were starting over. But no. You noticed that пѓќn, 1пѓћ пЂ« пѓќp, 1пѓћ пЂЅ пѓќn пЂ« p, 1пѓћ and
also пѓќn, 1пѓћпѓќp, 1пѓћ пЂЅ пѓќnp, 1пѓћ. Thus the set of all rational numbers whose second coordinate is
one behave just like the integers. If we simply abbreviate the rational number пѓќn, 1пѓћ by n,
there is absolutely no danger of confusion: 2 пЂ« 3 пЂЅ 5 stands for пѓќ2, 1пѓћ пЂ« пѓќ3, 1пѓћ пЂЅ пѓќ5, 1пѓћ. The
equation 3x пЂЅ 8 that started this all may then be interpreted as shorthand for the equation
пѓќ3, 1пѓћпѓќu, vпѓћ пЂЅ пѓќ8, 1пѓћ, and one easily verifies that x пЂЅ пѓќu, vпѓћ пЂЅ пѓќ8, 3пѓћ is a solution. Now, if
someone runs at you in the night and hands you a note with 5 written on it, you do not
know whether this is simply the integer 5 or whether it is shorthand for the rational number
пѓќ5, 1пѓћ. What we see is that it really doesnвЂ™t matter. What we have вЂќreallyвЂќ done is
expanded the collection of integers to the collection of rational numbers. In other words,
we can think of the set of all rational numbers as including the integersвЂ“they are simply the
rationals with second coordinate 1.

One last observation about rational numbers. It is, as everyone must know, traditional to

1.1
n n
write the ordered pair пѓќn, mпѓћ as . Thus n stands simply for the rational number , etc.
m 1

Now why have we spent this time on something everyone learned in the second grade?
Because this is almost a paradigm for what we do in constructing or defining the so-called
complex numbers. Watch.

Euclid showed us there is no rational solution to the equation x 2 пЂЅ 2. We were thus led to
defining even more new numbers, the so-called real numbers, which, of course, include the
rationals. This is hard, and you likely did not see it done in elementary school, but we shall
assume you know all about it and move along to the equation x 2 пЂЅ пЂї1. Now we define
complex numbers. These are simply ordered pairs пѓќx, yпѓћ of real numbers, just as the
rationals are ordered pairs of integers. Two complex numbers are equal only when there
are actually the sameвЂ“that is пѓќx, yпѓћ пЂЅ пѓќu, vпѓћ precisely when x пЂЅ u and y пЂЅ v. We define the
sum and product of two complex numbers:

пѓќx, yпѓћ пЂ« пѓќu, vпѓћ пЂЅ пѓќx пЂ« u, y пЂ« vпѓћ

and
пѓќx, yпѓћпѓќu, vпѓћ пЂЅ пѓќxu пЂї yv, xv пЂ« yuпѓћ

As always, subtraction and division are the inverses of these operations.

Now letвЂ™s consider the arithmetic of the complex numbers with second coordinate 0:

пѓќx, 0пѓћ пЂ« пѓќu, 0пѓћ пЂЅ пѓќx пЂ« u, 0пѓћ,

and
пѓќx, 0пѓћпѓќu, 0пѓћ пЂЅ пѓќxu, 0пѓћ.

Note that what happens is completely analogous to what happens with rationals with
second coordinate 1. We simply use x as an abbreviation for пѓќx, 0пѓћ and there is no danger of
confusion: x пЂ« u is short-hand for пѓќx, 0пѓћ пЂ« пѓќu, 0пѓћ пЂЅ пѓќx пЂ« u, 0пѓћ and xu is short-hand for
пѓќx, 0пѓћпѓќu, 0пѓћ. We see that our new complex numbers include a copy of the real numbers, just
as the rational numbers include a copy of the integers.

Next, notice that xпѓќu, vпѓћ пЂЅ пѓќu, vпѓћx пЂЅ пѓќx, 0пѓћпѓќu, vпѓћ пЂЅ пѓќxu, xvпѓћ. Now then, any complex number
z пЂЅ пѓќx, yпѓћ may be written

1.2
z пЂЅ пѓќx, yпѓћ пЂЅ пѓќx, 0пѓћ пЂ« пѓќ0, yпѓћ
пЂЅ x пЂ« yпѓќ0, 1пѓћ

When we let пЃЉ пЂЅ пѓќ0, 1пѓћ, then we have

z пЂЅ пѓќx, yпѓћ пЂЅ x пЂ« пЃЉy

Now, suppose z пЂЅ пѓќx, yпѓћ пЂЅ x пЂ« пЃЉy and w пЂЅ пѓќu, vпѓћ пЂЅ u пЂ« пЃЉv. Then we have

zw пЂЅ пѓќx пЂ« пЃЉyпѓћпѓќu пЂ« пЃЉvпѓћ
пЂЅ xu пЂ« пЃЉпѓќxv пЂ« yuпѓћ пЂ« пЃЉ 2 yv

We need only see what пЃЉ 2 is: пЃЉ 2 пЂЅ пѓќ0, 1пѓћпѓќ0, 1пѓћ пЂЅ пѓќпЂї1, 0пѓћ, and we have agreed that we can
safely abbreviate пѓќпЂї1, 0пѓћ as пЂї1. Thus, пЃЉ 2 пЂЅ пЂї1, and so

zw пЂЅ пѓќxu пЂї yvпѓћ пЂ« пЃЉпѓќxv пЂ« yuпѓћ

and we have reduced the fairly complicated definition of complex arithmetic simply to
ordinary real arithmetic together with the fact that пЃЉ 2 пЂЅ пЂї1.

z
LetвЂ™s take a look at divisionвЂ“the inverse of multiplication. Thus stands for that complex
w
number you must multiply w by in order to get z . An example:

z пЂЅ x пЂ« пЃЉy пЂЅ x пЂ« пЃЉy пЂ¶ u пЂї пЃЉv
u пЂ« пЃЉv u пЂї пЃЉv
w u пЂ« пЃЉv
пѓќxu пЂ« yvпѓћ пЂ« пЃЉпѓќyu пЂї xvпѓћ
пЂЅ
u2 пЂ« v2
xu пЂ« yv yu пЂї xv
пЂЅ2 пЂ«пЃЉ 2
u пЂ« v2 u пЂ« v2

Note this is just fine except when u 2 пЂ« v 2 пЂЅ 0; that is, when u пЂЅ v пЂЅ 0. We may thus divide
by any complex number except 0 пЂЅ пѓќ0, 0пѓћ.

One final note in all this. Almost everyone in the world except an electrical engineer uses
the letter i to denote the complex number we have called пЃЉ. We shall accordingly use i
rather than пЃЉ to stand for the number пѓќ0, 1пѓћ.

Exercises

1.3
1. Find the following complex numbers in the form x пЂ« iy:
b) пѓќ1 пЂї iпѓћ 3
a) пѓќ4 пЂї 7iпѓћпѓќпЂї2 пЂ« 3iпѓћ
b) пѓќ5пЂ«2iпѓћ c) 1i
пѓќ1пЂ«iпѓћ

2. Find all complex z пЂЅ пѓќx, yпѓћ such that
z2 пЂ« z пЂ« 1 пЂЅ 0

3. Prove that if wz пЂЅ 0, then w пЂЅ 0 or z пЂЅ 0.

1.2. Geometry. We now have this collection of all ordered pairs of real numbers, and so
there is an uncontrollable urge to plot them on the usual coordinate axes. We see at once
then there is a one-to-one correspondence between the complex numbers and the points in
the plane. In the usual way, we can think of the sum of two complex numbers, the point in
the plane corresponding to z пЂ« w is the diagonal of the parallelogram having z and w as
sides:

We shall postpone until the next section the geometric interpretation of the product of two
complex numbers.

The modulus of a complex number z пЂЅ x пЂ« iy is defined to be the nonnegative real number
x 2 пЂ« y 2 , which is, of course, the length of the vector interpretation of z. This modulus is
traditionally denoted |z|, and is sometimes called the length of z. Note that
|пѓќx, 0пѓћ| пЂЅ x 2 пЂЅ |x|, and so |пЂ¶| is an excellent choice of notation for the modulus.

The conjugate z of a complex number z пЂЅ x пЂ« iy is defined by z пЂЅ x пЂї iy. Thus |z| 2 пЂЅ z z .
Geometrically, the conjugate of z is simply the reflection of z in the horizontal axis:

1.4
Observe that if z пЂЅ x пЂ« iy and w пЂЅ u пЂ« iv, then

пѓќz пЂ« wпѓћ пЂЅ пѓќx пЂ« uпѓћ пЂї iпѓќy пЂ« vпѓћ
пЂЅ пѓќx пЂї iyпѓћ пЂ« пѓќu пЂї ivпѓћ
пЂЅ z пЂ« w.

In other words, the conjugate of the sum is the sum of the conjugates. It is also true that
zw пЂЅ z w. If z пЂЅ x пЂ« iy, then x is called the real part of z, and y is called the imaginary
part of z. These are usually denoted Re z and Im z, respectively. Observe then that
z пЂ« z пЂЅ 2 Re z and z пЂї z пЂЅ 2 Im z.

Now, for any two complex numbers z and w consider

|z пЂ« w| 2 пЂЅ пѓќz пЂ« wпѓћпѓќz пЂ« wпѓћ пЂЅ пѓќz пЂ« wпѓћпѓќ z пЂ« wпѓћ
пЂЅ z z пЂ« пѓќw z пЂ« wzпѓћ пЂ« ww
пЂЅ |z| 2 пЂ« 2 Reпѓќw z пѓћ пЂ« |w| 2
п‚І |z| 2 пЂ« 2|z||w| пЂ« |w| 2 пЂЅ пѓќ|z| пЂ« |w|пѓћ 2
In other words,
|z пЂ« w| п‚І |z| пЂ« |w|
the so-called triangle inequality. (This inequality is an obvious geometric factвЂ“can you
guess why it is called the triangle inequality?)

Exercises

4. a)Prove that for any two complex numbers, zw пЂЅ z w.
z z
b)Prove that пѓќ w пѓћ пЂЅ w .
c)Prove that ||z| пЂї |w|| п‚І |z пЂї w|.

|z|
z
5. Prove that |zw| пЂЅ |z||w| and that | w | пЂЅ .
|w|

1.5
6. Sketch the set of points satisfying
a) |z пЂї 2 пЂ« 3i| пЂЅ 2 b)|z пЂ« 2i| п‚І 1
c) Reпѓќ z пЂ« iпѓћ пЂЅ 4 d) |z пЂї 1 пЂ« 2i| пЂЅ |z пЂ« 3 пЂ« i|
e)|z пЂ« 1| пЂ« |z пЂї 1| пЂЅ 4 f) |z пЂ« 1| пЂї |z пЂї 1| пЂЅ 4

1.3. Polar coordinates. Now letвЂ™s look at polar coordinates пѓќr, пЃ“пѓћ of complex numbers.
Then we may write z пЂЅ rпѓќcos пЃ“ пЂ« i sin пЃ“пѓћ. In complex analysis, we do not allow r to be
negative; thus r is simply the modulus of z. The number пЃ“ is called an argument of z, and
there are, of course, many different possibilities for пЃ“. Thus a complex numbers has an
infinite number of arguments, any two of which differ by an integral multiple of 2пЃћ. We
usually write пЃ“ пЂЅ arg z. The principal argument of z is the unique argument that lies on
the interval пѓќпЂїпЃћ, пЃћпѓ .

Example. For 1 пЂї i, we have

2 пѓќcos 7пЃћ пЂ« i sin 7пЃћ пѓћ
1пЂїi пЂЅ
4 4
2 пѓќcos пЂї пЃћ пЂ« i sin пЂї пЃћ пѓћ
пЂЅ
4 4
2 пѓќcos 399пЃћ пЂ« i sin 399пЃћ пѓћ
пЂЅ
4 4

, пЂї пЃћ , and
7пЃћ 399пЃћ
is an argument of 1 пЂї i, but the
etc., etc., etc. Each of the numbers 4 4 4
principal argument is пЂї пЃћ .
4

Suppose z пЂЅ rпѓќcos пЃ“ пЂ« i sin пЃ“пѓћ and w пЂЅ sпѓќcos пЃ™ пЂ« i sin пЃ™пѓћ. Then
zw пЂЅ rпѓќcos пЃ“ пЂ« i sin пЃ“пѓћsпѓќcos пЃ™ пЂ« i sin пЃ™пѓћ
пЂЅ rsпѓџпѓќcos пЃ“ cos пЃ™ пЂї sin пЃ“ sin пЃ™пѓћ пЂ« iпѓќsin пЃ“ cos пЃ™ пЂ« sin пЃ™ cos пЃ“пѓћ пѓ
пЂЅ rsпѓќcosпѓќпЃ“ пЂ« пЃ™пѓћ пЂ« i sinпѓќпЃ“ пЂ« пЃ™пѓћ пѓћ
We have the nice result that the product of two complex numbers is the complex number
whose modulus is the product of the moduli of the two factors and an argument is the sum
of arguments of the factors. A picture:

1.6
We now define expпѓќiпЃ“пѓћ, or e iпЃ“ by
e iпЃ“ пЂЅ cos пЃ“ пЂ« i sin пЃ“

We shall see later as the drama of the term unfolds that this very suggestive notation is an
excellent choice. Now, we have in polar form

z пЂЅ re iпЃ“ ,

where r пЂЅ |z| and пЃ“ is any argument of z. Observe we have just shown that

e iпЃ“ e iпЃ™ пЂЅ e iпѓќпЃ“пЂ«пЃ™пѓћ .

It follows from this that e iпЃ“ e пЂїiпЃ“ пЂЅ 1. Thus

1 пЂЅ e пЂїiпЃ“
e iпЃ“

It is easy to see that

z пЂЅ re iпЃ“ пЂЅ r пѓќcosпѓќпЃ“ пЂї пЃ™пѓћ пЂ« i sinпѓќпЃ“ пЂї пЃ™пѓћпѓћ
w s
se iпЃ™

Exercises

7. Write in polar form re iпЃ“ :
b) 1 пЂ« i
a) i
c) пЂї2 d) пЂї3i
e) 3 пЂ« 3i

8. Write in rectangular formвЂ”no decimal approximations, no trig functions:
a) 2e i3пЃћ b) e i100пЃћ
c) 10e iпЃћ/6 d) 2 e i5пЃћ/4

9. a) Find a polar form of пѓќ1 пЂ« iпѓћпѓќ1 пЂ« i 3 пѓћ.
b) Use the result of a) to find cos 7пЃћ and sin 7пЃћ
.
12 12

10. Find the rectangular form of пѓќпЂї1 пЂ« iпѓћ 100 .

1.7
11. Find all z such that z 3 пЂЅ 1. (Again, rectangular form, no trig functions.)

12. Find all z such that z 4 пЂЅ 16i. (Rectangular form, etc.)

1.8
Chapter Two

Complex Functions

2.1. Functions of a real variable. A function пЃЊ : I п‚ё C from a set I of reals into the
complex numbers C is actually a familiar concept from elementary calculus. It is simply a
function from a subset of the reals into the plane, what we sometimes call a vector-valued
function. Assuming the function пЃЊ is nice, it provides a vector, or parametric, description
of a curve. Thus, the set of all пѓЎпЃЊпѓќtпѓћ : пЃЊпѓќtпѓћ пЂЅ e it пЂЅ cos t пЂ« i sin t пЂЅ пѓќcos t, sin tпѓћ, 0 п‚І t п‚І 2пЃћпѓў
is the circle of radius one, centered at the origin.

We also already know about the derivatives of such functions. If пЃЊпѓќtпѓћ пЂЅ xпѓќtпѓћ пЂ« iyпѓќtпѓћ, then
the derivative of пЃЊ is simply пЃЊ пЃ¶ пѓќtпѓћ пЂЅ x пЃ¶ пѓќtпѓћ пЂ« iy пЃ¶ пѓќtпѓћ, interpreted as a vector in the plane, it is
tangent to the curve described by пЃЊ at the point пЃЊпѓќtпѓћ.

Example. Let пЃЊпѓќtпѓћ пЂЅ t пЂ« it 2 , пЂї1 п‚І t п‚І 1. One easily sees that this function describes that
part of the curve y пЂЅ x 2 between x пЂЅ пЂї1 and x пЂЅ 1:

1

0
-1 -0.5 0.5 1
x

Another example. Suppose there is a body of mass M вЂќfixedвЂќ at the originвЂ“perhaps the
sunвЂ“and there is a body of mass m which is free to moveвЂ“perhaps a planet. Let the location
of this second body at time t be given by the complex-valued function zпѓќtпѓћ. We assume the
only force on this mass is the gravitational force of the fixed body. This force f is thus

zпѓќtпѓћ
f пЂЅ GMm пЂї
|zпѓќtпѓћ| 2 |zпѓќtпѓћ|

where G is the universal gravitational constant. Sir Isaac Newton tells us that

zпѓќtпѓћ
mz пЃ¶пЃ¶ пѓќtпѓћ пЂЅ f пЂЅ GMm пЂї
|zпѓќtпѓћ| 2 |zпѓќtпѓћ|

2.1
Hence,

z пЃ¶пЃ¶ пЂЅ пЂї GM z
|z| 3

Next, letвЂ™s write this in polar form, z пЂЅ re iпЃ“ :

d 2 пѓќre iпЃ“ пѓћ пЂЅ пЂї k e iпЃ“
dt 2 r2

where we have written GM пЂЅ k. Now, letвЂ™s see what we have.

d пѓќre iпЃ“ пѓћ пЂЅ r d пѓќe iпЃ“пѓћ пѓћ пЂ« dr e iпЃ“
dt dt dt

Now,

d пѓќe iпЃ“пѓћ пѓћ пЂЅ d пѓќcos пЃ“ пЂ« i sin пЃ“ пѓћ
dt dt
пЂЅ пѓќпЂї sin пЃ“ пЂ« i cos пЃ“ пѓћ dпЃ“
dt
пЂЅ iпѓќcos пЃ“ пЂ« i sin пЃ“пѓћ dпЃ“
dt
пЂЅ i dпЃ“ e iпЃ“ .
dt

(Additional evidence that our notation e iпЃ“ пЂЅ cos пЃ“ пЂ« i sin пЃ“ is reasonable.)
Thus,

d пѓќre iпЃ“ пѓћ пЂЅ r d пѓќe iпЃ“пѓћ пѓћ пЂ« dr e iпЃ“
dt dt dt
пЂЅ r i dпЃ“ e iпЃ“ пЂ« dr e iпЃ“
dt dt
пЂЅ dr пЂ« ir dпЃ“ e iпЃ“ .
dt dt

Now,

2.2
dпЃ“ пЂ« ir d 2 пЃ“ e iпЃ“ пЂ«
d 2 пѓќre iпЃ“ пѓћ пЂЅ d 2 r пЂ« i dr
dt dt
dt 2 dt 2 dt 2
dr пЂ« ir dпЃ“ i dпЃ“ e iпЃ“
dt dt dt
dпЃ“ 2 пЂ« i r d 2 пЃ“ пЂ« 2 dr dпЃ“
d2r пЂї r e iпЃ“
пЂЅ
dt dt dt
dt 2 dt 2

d2
пѓќre iпЃ“ пѓћ пЂЅ пЂї rk2 e iпЃ“ becomes
Now, the equation dt 2

2 2
d 2 r пЂї r dпЃ“ пЃ“
пЂ« i r d 2 пЂ« 2 dr dпЃ“ пЂЅ пЂї k2 .
dt dt dt
dt 2 dt r

This gives us the two equations

2
d 2 r пЂї r dпЃ“ пЂЅ пЂї k2 ,
dt
dt 2 r

and,

2
пЃ“
r d 2 пЂ« 2 dr dпЃ“ пЂЅ 0.
dt dt
dt

Multiply by r and this second equation becomes

d r 2 dпЃ“ пЂЅ 0.
dt dt

This tells us that
пЃЉ пЂЅ r 2 dпЃ“
dt

is a constant. (This constant пЃЉ is called the angular momentum.) This result allows us to
get rid of dпЃ“ in the first of the two differential equations above:
dt

2
d2r пЂї r пЃЉ пЂЅ пЂї k2
dt 2 r2 r
or,

d2r пЂї пЃЉ2 пЂЅ пЂї k .
dt 2 r3 r2

2.3
Although this now involves only the one unknown function r, as it stands it is tough to
solve. LetвЂ™s change variables and think of r as a function of пЃ“. LetвЂ™s also write things in
terms of the function s пЂЅ 1 . Then,
r

d пЂЅ dпЃ“ d пЂЅ пЃЉ d .
dt dt dпЃ“ r 2 dпЃ“

Hence,
dr пЂЅ пЃЉ dr пЂЅ пЂїпЃЉ ds ,
dt r 2 dпЃ“ dпЃ“
and so
d 2 r пЂЅ d пЂїпЃЉ ds пЂЅ пЃЉs 2 d пЂїпЃЉ ds
dt dпЃ“ dпЃ“ dпЃ“
dt 2
2
пЂЅ пЂїпЃЉ 2 s 2 d s ,
dпЃ“ 2

and our differential equation looks like

d 2 r пЂї пЃЉ 2 пЂЅ пЂїпЃЉ 2 s 2 d 2 s пЂї пЃЉ 2 s 3 пЂЅ пЂїks 2 ,
dt 2 r3 dпЃ“ 2
or,
d2s пЂ« s пЂЅ k .
dпЃ“ 2 пЃЉ2

This one is easy. From high school differential equations class, we remember that

s пЂЅ 1 пЂЅ A cosпѓќпЃ“ пЂ« пЃЄпѓћ пЂ« k2 ,
r пЃЉ

where A and пЃЄ are constants which depend on the initial conditions. At long last,

пЃЉ 2 /k
rпЂЅ ,
1 пЂ« пЃђ cosпѓќпЃ“ пЂ« пЃЄпѓћ

where we have set пЃђ пЂЅ AпЃЉ 2 /k. The graph of this equation is, of course, a conic section of
eccentricity пЃђ.

Exercises

2.4
1. a)What curve is described by the function пЃЊпѓќtпѓћ пЂЅ пѓќ3t пЂ« 4пѓћ пЂ« iпѓќt пЂї 6пѓћ, 0 п‚І t п‚І 1 ?
b)Suppose z and w are complex numbers. What is the curve described by
пЃЊпѓќtпѓћ пЂЅ пѓќ1 пЂї tпѓћw пЂ« tz, 0 п‚І t п‚І 1 ?

2. Find a function пЃЊ that describes that part of the curve y пЂЅ 4x 3 пЂ« 1 between x пЂЅ 0 and
x пЂЅ 10.

3. Find a function пЃЊ that describes the circle of radius 2 centered at z пЂЅ 3 пЂї 2i .

4. Note that in the discussion of the motion of a body in a central gravitational force field,
it was assumed that the angular momentum пЃЉ is nonzero. Explain what happens in case
пЃЉ пЂЅ 0.

2.2 Functions of a complex variable. The real excitement begins when we consider
function f : D п‚ё C in which the domain D is a subset of the complex numbers. In some
sense, these too are familiar to us from elementary calculusвЂ”they are simply functions
from a subset of the plane into the plane:

fпѓќzпѓћ пЂЅ fпѓќx, yпѓћ пЂЅ uпѓќx, yпѓћ пЂ« ivпѓќx, yпѓћ пЂЅ пѓќuпѓќx, yпѓћ, vпѓќx, yпѓћпѓћ

Thus fпѓќzпѓћ пЂЅ z 2 looks like fпѓќzпѓћ пЂЅ z 2 пЂЅ пѓќx пЂ« iyпѓћ 2 пЂЅ x 2 пЂї y 2 пЂ« 2xyi. In other words,
uпѓќx, yпѓћ пЂЅ x 2 пЂї y 2 and vпѓќx, yпѓћ пЂЅ 2xy. The complex perspective, as we shall see, generally
provides richer and more profitable insights into these functions.

The definition of the limit of a function f at a point z пЂЅ z 0 is essentially the same as that
which we learned in elementary calculus:

lim fпѓќzпѓћ пЂЅ L
zп‚ёz 0

means that given an пЃђ пЂѕ 0, there is a пЃЋ so that |fпѓќzпѓћ пЂї L| пЂј пЃђ whenever 0 пЂј |z пЂї z 0 | пЂј пЃЋ. As
you could guess, we say that f is continuous at z 0 if it is true that lim fпѓќzпѓћ пЂЅ fпѓќz 0 пѓћ. If f is
zп‚ёz 0
continuous at each point of its domain, we say simply that f is continuous.

Suppose both lim fпѓќzпѓћ and lim gпѓќzпѓћ exist. Then the following properties are easy to
zп‚ёz 0 zп‚ёz 0
establish:

2.5
lim пѓџfпѓќzпѓћ п‚± gпѓќzпѓћ пѓ  пЂЅlim fпѓќzпѓћ п‚±lim gпѓќzпѓћ
zп‚ёz 0 zп‚ёz 0 zп‚ёz 0

lim fпѓќzпѓћgпѓќzпѓћ пЂЅlim fпѓќzпѓћ lim gпѓќzпѓћ
zп‚ёz 0 zп‚ёz 0 zп‚ёz 0

and,
lim fпѓќzпѓћ
fпѓќzпѓћ
пЂЅ zп‚ёz 0
lim
zп‚ёz 0 gпѓќzпѓћ lim gпѓќzпѓћ
zп‚ёz 0

provided, of course, that lim gпѓќzпѓћ п‚® 0.
zп‚ёz 0

It now follows at once from these properties that the sum, difference, product, and quotient
of two functions continuous at z 0 are also continuous at z 0 . (We must, as usual, except the

It should not be too difficult to convince yourself that if z пЂЅ пѓќx, yпѓћ, z 0 пЂЅ пѓќx 0 , y 0 пѓћ, and
fпѓќzпѓћ пЂЅ uпѓќx, yпѓћ пЂ« ivпѓќx, yпѓћ, then

lim fпѓќzпѓћ пЂЅ uпѓќx, yпѓћ пЂ« i
lim lim vпѓќx, yпѓћ
zп‚ёz 0 пѓќx,yпѓћп‚ёпѓќx 0 ,y 0 пѓћ пѓќx,yпѓћп‚ёпѓќx 0 ,y 0 пѓћ

Thus f is continuous at z 0 пЂЅ пѓќx 0 , y 0 пѓћ precisely when u and v are.

Our next step is the definition of the derivative of a complex function f. It is the obvious
thing. Suppose f is a function and z 0 is an interior point of the domain of f . The derivative
f пЃ¶ пѓќz 0 пѓћ of f is

fпѓќzпѓћ пЂї fпѓќz 0 пѓћ
f пЃ¶ пѓќz 0 пѓћ пЂЅlim z пЂї z0
zп‚ёz 0

Example

Suppose fпѓќzпѓћ пЂЅ z 2 . Then, letting пЃЃz пЂЅ z пЂї z 0 , we have

2.6
fпѓќzпѓћ пЂї fпѓќz 0 пѓћ fпѓќz 0 пЂ« пЃЃzпѓћ пЂї fпѓќz 0 пѓћ
пЂЅ lim
lim z пЂї z0 пЃЃz
zп‚ёz 0 пЃЃzп‚ё0
пѓќz 0 пЂ« пЃЃzпѓћ 2 пЂї z 2
0
пЂЅ lim
пЃЃz
пЃЃzп‚ё0
2z 0 пЃЃz пЂ« пѓќпЃЃzпѓћ 2
пЂЅ lim
пЃЃz
пЃЃzп‚ё0

пЂЅ lim пѓќ2z 0 пЂ« пЃЃz пѓћ
пЃЃzп‚ё0

пЂЅ 2z 0

No surprise hereвЂ“the function fпѓќzпѓћ пЂЅ z 2 has a derivative at every z, and itвЂ™s simply 2z.

Another Example

Let fпѓќzпѓћ пЂЅ zz. Then,

fпѓќz 0 пЂ« пЃЃzпѓћ пЂї fпѓќz 0 пѓћ пѓќz 0 пЂ« пЃЃzпѓћпѓќz 0 пЂ« пЃЃzпѓћ пЂї z 0 z 0
пЂЅ lim
lim
пЃЃz пЃЃz
пЃЃzп‚ё0 пЃЃzп‚ё0

пЂЅ lim z 0 пЃЃz пЂ« z 0 пЃЃz пЂ« пЃЃzпЃЃz
пЃЃz
пЃЃzп‚ё0

z 0 пЂ« пЃЃz пЂ« z 0 пЃЃz
пЂЅ lim
пЃЃz
пЃЃzп‚ё0

Suppose this limit exists, and choose пЃЃz пЂЅ пѓќпЃЃx, 0пѓћ. Then,

z 0 пЂ« пЃЃz пЂ« z 0 пЃЃz z 0 пЂ« пЃЃx пЂ« z 0 пЃЃx
пЂЅ lim
lim
пЃЃz пЃЃx
пЃЃzп‚ё0 пЃЃxп‚ё0

пЂЅ z 0 пЂ« z0
Now, choose пЃЃz пЂЅ пѓќ0, пЃЃyпѓћ. Then,

iпЃЃy
z 0 пЂ« пЃЃz пЂ« z 0 пЃЃz пЂЅ lim z 0 пЂї iпЃЃy пЂї z 0
lim
пЃЃz iпЃЃy
пЃЃyп‚ё0
пЃЃzп‚ё0

пЂЅ z 0 пЂї z0

Thus, we must have z 0 пЂ« z 0 пЂЅ z 0 пЂї z 0 , or z 0 пЂЅ 0. In other words, there is no chance of
this limitвЂ™s existing, except possibly at z 0 пЂЅ 0. So, this function does not have a derivative
at most places.

Now, take another look at the first of these two examples. It looks exactly like what you

2.7
did in Mrs. TurnerвЂ™s 3 rd grade calculus class for plain old real-valued functions. Meditate
on this and you will be convinced that all the вЂќusualвЂќ results for real-valued functions also
hold for these new complex functions: the derivative of a constant is zero, the derivative of
the sum of two functions is the sum of the derivatives, the вЂќproductвЂќ and вЂќquotientвЂќ rules
for derivatives are valid, the chain rule for the composition of functions holds, etc., etc. For
proofs, you need only go back to your elementary calculus book and change xвЂ™s to zвЂ™s.

A bit of jargon is in order. If f has a derivative at z 0 , we say that f is differentiable at z 0 . If
f is differentiable at every point of a neighborhood of z 0 , we say that f is analytic at z 0 . (A
set S is a neighborhood of z 0 if there is a disk D пЂЅ пѓЎz : |z пЂї z 0 | пЂј r, r пЂѕ 0 пѓў so that D пѓђ S.
) 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.

Exercises

5. Suppose fпѓќzпѓћ пЂЅ 3xy пЂ« iпѓќx пЂї y 2 пѓћ. Find lim fпѓќzпѓћ, or explain carefully why it does not
zп‚ё3пЂ«2i
exist.

6. Prove that if f has a derivative at z, then f is continuous at z.

7. Find all points at which the valued function f defined by fпѓќzпѓћ пЂЅ z has a derivative.

8. Find all points at which the valued function f defined by
fпѓќzпѓћ пЂЅ пѓќ2 пЂ« iпѓћz 3 пЂї iz 2 пЂ« 4z пЂї пѓќ1 пЂ« 7iпѓћ
has a derivative.

9. Is the function f given by
пѓќ z пѓћ2
,z п‚® 0
z
fпѓќzпѓћ пЂЅ
,z пЂЅ 0
0

differentiable at z пЂЅ 0? Explain.

2.3. Derivatives. Suppose the function f given by fпѓќzпѓћ пЂЅ uпѓќx, yпѓћ пЂ« ivпѓќx, yпѓћ has a derivative
at z пЂЅ z 0 пЂЅ пѓќx 0 , y 0 пѓћ. We know this means there is a number f пЃ¶ пѓќz 0 пѓћ so that

fпѓќz 0 пЂ« пЃЃzпѓћ пЂї fпѓќz 0 пѓћ
f пЃ¶ пѓќz 0 пѓћ пЂЅ lim .
пЃЃz
пЃЃzп‚ё0

2.8
Choose пЃЃz пЂЅ пѓќпЃЃx, 0пѓћ пЂЅ пЃЃx. Then,

fпѓќz 0 пЂ« пЃЃzпѓћ пЂї fпѓќz 0 пѓћ
f пЃ¶ пѓќz 0 пѓћ пЂЅ lim
пЃЃz
пЃЃzп‚ё0
uпѓќx 0 пЂ« пЃЃx, y 0 пѓћ пЂ« ivпѓќx 0 пЂ« пЃЃx, y 0 пѓћ пЂї uпѓќx 0 , y 0 пѓћ пЂї ivпѓќx 0 , y 0 пѓћ
пЂЅ lim
пЃЃx
пЃЃxп‚ё0
uпѓќx 0 пЂ« пЃЃx, y 0 пѓћ пЂї uпѓќx 0 , y 0 пѓћ vпѓќx 0 пЂ« пЃЃx, y 0 пѓћ пЂї vпѓќx 0 , y 0 пѓћ
пЂЅ lim пЂ«i
пЃЃx пЃЃx
пЃЃxп‚ё0

пЂЅ пЂЇu пѓќx 0 , y 0 пѓћ пЂ« i пЂЇv пѓќx 0 , y 0 пѓћ
пЂЇx пЂЇx

Next, choose пЃЃz пЂЅ пѓќ0, пЃЃyпѓћ пЂЅ iпЃЃy. Then,

fпѓќz 0 пЂ« пЃЃzпѓћ пЂї fпѓќz 0 пѓћ
f пЃ¶ пѓќz 0 пѓћ пЂЅ lim
пЃЃz
пЃЃzп‚ё0
uпѓќx 0 , y 0 пЂ« пЃЃyпѓћ пЂ« ivпѓќx 0 , y 0 пЂ« пЃЃyпѓћ пЂї uпѓќx 0 , y 0 пѓћ пЂї ivпѓќx 0 , y 0 пѓћ
пЂЅ lim
iпЃЃy
пЃЃyп‚ё0
vпѓќx 0 , y 0 пЂ« пЃЃyпѓћ пЂї vпѓќx 0 , y 0 пѓћ uпѓќx 0 , y 0 пЂ« пЃЃyпѓћ пЂї uпѓќx 0 , y 0 пѓћ
пЂЅ lim пЂїi
пЃЃy пЃЃy
пЃЃyп‚ё0

пЂЅ пЂЇv пѓќx 0 , y 0 пѓћ пЂї i пЂЇu пѓќx 0 , y 0 пѓћ
пЂЇy пЂЇy

We have two different expressions for the derivative f пЃ¶ пѓќz 0 пѓћ, and so

пЂЇu пѓќx 0 , y 0 пѓћ пЂ« i пЂЇv пѓќx 0 , y 0 пѓћ пЂЅ пЂЇv пѓќx 0 , y 0 пѓћ пЂї i пЂЇu пѓќx 0 , y 0 пѓћ
пЂЇx пЂЇx пЂЇy пЂЇy
or,

пЂЇu пѓќx 0 , y 0 пѓћ пЂЅ пЂЇv пѓќx 0 , y 0 пѓћ,
пЂЇx пЂЇy
пЂЇu пѓќx 0 , y 0 пѓћ пЂЅ пЂї пЂЇv пѓќx 0 , y 0 пѓћ
пЂЇy пЂЇx

These equations are called the Cauchy-Riemann Equations.

We have shown that if f has a derivative at a point z 0 , 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 uпѓќx, yпѓћ and vпѓќx, yпѓћ have partial
derivatives in a neighborhood of z 0 пЂЅ пѓќx 0 , y 0 пѓћ, suppose these derivatives are continuous at
z 0 , and suppose

2.9
пЂЇu пѓќx 0 , y 0 пѓћ пЂЅ пЂЇv пѓќx 0 , y 0 пѓћ,
пЂЇx пЂЇy
пЂЇu пѓќx 0 , y 0 пѓћ пЂЅ пЂї пЂЇv пѓќx 0 , y 0 пѓћ.
пЂЇy пЂЇx

We shall see that f is differentiable at z 0 .

fпѓќz 0 пЂ« пЃЃzпѓћ пЂї fпѓќz 0 пѓћ
пЃЃz
пѓџuпѓќx 0 пЂ« пЃЃx, y 0 пЂ« пЃЃyпѓћ пЂї uпѓќx 0 , y 0 пѓћ пѓ  пЂ« iпѓџvпѓќx 0 пЂ« пЃЃx, y 0 пЂ« пЃЃyпѓћ пЂї vпѓќx 0 , y 0 пѓћ пѓ
пЂЅ .
пЃЃx пЂ« iпЃЃy

Observe that

uпѓќx 0 пЂ« пЃЃx, y 0 пЂ« пЃЃyпѓћ пЂї uпѓќx 0 , y 0 пѓћ пЂЅ пѓџuпѓќx 0 пЂ« пЃЃx, y 0 пЂ« пЃЃyпѓћ пЂї uпѓќx 0 , y 0 пЂ« пЃЃyпѓћ пѓ  пЂ«
пѓџuпѓќx 0 , y 0 пЂ« пЃЃyпѓћ пЂї uпѓќx 0 , y 0 пѓћ пѓ .

Thus,
uпѓќx 0 пЂ« пЃЃx, y 0 пЂ« пЃЃyпѓћ пЂї uпѓќx 0 , y 0 пЂ« пЃЃyпѓћ пЂЅ пЃЃx пЂЇu пѓќпЃ™, y 0 пЂ« пЃЃyпѓћ,
пЂЇx

and,
пЂЇu пѓќпЃ™, y пЂ« пЃЃyпѓћ пЂЅ пЂЇu пѓќx , y пѓћ пЂ« пЃђ ,
0 00 1
пЂЇx пЂЇx
where,
lim пЃђ 1 пЂЅ 0.
пЃЃzп‚ё0

Thus,
uпѓќx 0 пЂ« пЃЃx, y 0 пЂ« пЃЃyпѓћ пЂї uпѓќx 0 , y 0 пЂ« пЃЃyпѓћ пЂЅ пЃЃx пЂЇu пѓќx 0 , y 0 пѓћ пЂ« пЃђ 1 .
пЂЇx

Proceeding similarly, we get

2.10
fпѓќz 0 пЂ« пЃЃzпѓћ пЂї fпѓќz 0 пѓћ
пЃЃz
пѓџuпѓќx 0 пЂ« пЃЃx, y 0 пЂ« пЃЃyпѓћ пЂї uпѓќx 0 , y 0 пѓћ пѓ  пЂ« iпѓџvпѓќx 0 пЂ« пЃЃx, y 0 пЂ« пЃЃyпѓћ пЂї vпѓќx 0 , y 0 пѓћ пѓ
пЂЅ
пЃЃx пЂ« iпЃЃy
пЂЇu
пѓќx 0 , y 0 пѓћ пЂ« пЃђ 1 пЂ« i пЂЇv пѓќx 0 , y 0 пѓћ пЂ« iпЃђ 2 пЂ« пЃЃy пЂЇu
пѓќx 0 , y 0 пѓћ пЂ« пЃђ 3 пЂ« i пЂЇv пѓќx 0 , y 0 пѓћ пЂ« iпЃђ 4
пЃЃx пЂЇx пЂЇx пЂЇy пЂЇy
пЂЅ ,.
пЃЃx пЂ« iпЃЃy

where пЃђ i п‚ё 0 as пЃЃz п‚ё 0. Now, unleash the Cauchy-Riemann equations on this quotient and
obtain,

fпѓќz 0 пЂ« пЃЃzпѓћ пЂї fпѓќz 0 пѓћ
пЃЃz
пЃЃx пЂЇu пЂ« i пЂЇv пЂ« iпЃЃy пЂЇu пЂ« i пЂЇv stuff
пЂЇx пЂЇx пЂЇx пЂЇx
пЂЅ пЂ«
пЃЃx пЂ« iпЃЃy пЃЃx пЂ« iпЃЃy
пЂЅ пЂЇu пЂ« i пЂЇv пЂ« stuff .
пЃЃx пЂ« iпЃЃy
пЂЇx пЂЇx
Here,
stuff пЂЅ пЃЃxпѓќпЃђ 1 пЂ« iпЃђ 2 пѓћ пЂ« пЃЃyпѓќпЃђ 3 пЂ« iпЃђ 4 пѓћ.

ItвЂ™s easy to show that

lim stuff пЂЅ 0,
пЃЃz
пЃЃzп‚ё0

and so,
fпѓќz 0 пЂ« пЃЃzпѓћ пЂї fпѓќz 0 пѓћ
пЂЅ пЂЇu пЂ« i пЂЇv .
lim
пЃЃz пЂЇx пЂЇx
пЃЃzп‚ё0

In particular we have, as promised, shown that f is differentiable at z 0 .

Example

LetвЂ™s find all points at which the function f given by fпѓќzпѓћ пЂЅ x 3 пЂ« iпѓќ1 пЂї yпѓћ 3 is differentiable.
Here we have u пЂЅ x 3 and v пЂЅ пѓќ1 пЂї yпѓћ 3 . The Cauchy-Riemann equations thus look like

3x 2 пЂЅ 3пѓќ1 пЂї yпѓћ 2 , and
0 пЂЅ 0.

2.11
The partial derivatives of u and v are nice and continuous everywhere, so f will be
differentiable everywhere the C-R equations are satisfied. That is, everywhere

x 2 пЂЅ пѓќ1 пЂї yпѓћ 2 ; that is, where
x пЂЅ 1 пЂї y, or x пЂЅ пЂї1 пЂ« y.

This is simply the set of all points on the cross formed by the two straight lines

4

3

2

1

0
-3 -2 -1 1 2 3
x
-1

-2

Exercises

10. At what points is the function f given by fпѓќzпѓћ пЂЅ x 3 пЂ« iпѓќ1 пЂї yпѓћ 3 analytic? Explain.

11. Do the real and imaginary parts of the function f in Exercise 9 satisfy the
Cauchy-Riemann equations at z пЂЅ 0? What do you make of your answer?

12. Find all points at which fпѓќzпѓћ пЂЅ 2y пЂї ix is differentiable.

13. Suppose f is analytic on a connected open set D, and f пЃ¶ пѓќzпѓћ пЂЅ 0 for all zпЃЏD. Prove that f
is constant.

14. Find all points at which
y
x
fпѓќzпѓћ пЂЅ пЂїi 2
x2 пЂ« y2 x пЂ« y2

is differentiable. At what points is f analytic? Explain.

15. Suppose f is analytic on the set D, and suppose Re f is constant on D. Is f necessarily

2.12
constant on D? Explain.

16. Suppose f is analytic on the set D, and suppose |fпѓќzпѓћ| is constant on D. Is f necessarily
constant on D? Explain.

2.13
Chapter Three

Elementary Functions

3.1. Introduction. Complex functions are, of course, quite easy to come byвЂ”they are
simply ordered pairs of real-valued functions of two variables. We have, however, already
seen enough to realize that it is those complex functions that are differentiable that are the
most interesting. It was important in our invention of the complex numbers that these new
numbers in some sense included the old real numbersвЂ”in other words, we extended the
reals. We shall find it most useful and profitable to do a similar thing with many of the
familiar real functions. That is, we seek complex functions such that when restricted to the
reals are familiar real functions. As we have seen, the extension of polynomials and
rational functions to complex functions is easy; we simply change xвЂ™s to zвЂ™s. Thus, for
instance, the function f defined by

2
fпѓќzпѓћ пЂЅ z пЂ« z пЂ« 1
zпЂ«1

has a derivative at each point of its domain, and for z пЂЅ x пЂ« 0i, becomes a familiar real
rational function

2
fпѓќxпѓћ пЂЅ x пЂ« x пЂ« 1 .
xпЂ«1

What happens with the trigonometric functions, exponentials, logarithms, etc., is not so
obvious. Let us begin.

3.2. The exponential function. Let the so-called exponential function exp be defined by

expпѓќzпѓћ пЂЅ e x пѓќcos y пЂ« i sin yпѓћ,

where, as usual, z пЂЅ x пЂ« iy. From the Cauchy-Riemann equations, we see at once that this
function has a derivative every whereвЂ”it is an entire function. Moreover,

d expпѓќzпѓћ пЂЅ expпѓќzпѓћ.
dz

Note next that if z пЂЅ x пЂ« iy and w пЂЅ u пЂ« iv, then

3.1
expпѓќz пЂ« wпѓћ пЂЅ e xпЂ«u пѓџcosпѓќy пЂ« vпѓћ пЂ« i sinпѓќy пЂ« vпѓћ пѓ
пЂЅ e x e y пѓџcos y cos v пЂї sin y sin v пЂ« iпѓќsin y cos v пЂ« cos y sin vпѓћ пѓ
пЂЅ e x e y пѓќcos y пЂ« i sin yпѓћпѓќcos v пЂ« i sin vпѓћ
пЂЅ expпѓќzпѓћ expпѓќwпѓћ.

We thus use the quite reasonable notation e z пЂЅ expпѓќzпѓћ and observe that we have extended
the real exponential e x to the complex numbers.

Example

Recall from elementary circuit analysis that the relation between the voltage drop V and the
current flow I through a resistor is V пЂЅ RI, where R is the resistance. For an inductor, the
relation is V пЂЅ L dI , where L is the inductance; and for a capacitor, C dV пЂЅ I, where C is
dt dt
the capacitance. (The variable t is, of course, time.) Note that if V is sinusoidal with a
frequency пЃ§, then so also is I. Suppose then that V пЂЅ A sinпѓќпЃ§t пЂ« пЃЄпѓћ. We can write this as
V пЂЅ ImпѓќAe iпЃЄ e iпЃ§t пѓћ пЂЅ ImпѓќBe iпЃ§t пѓћ, where B is complex. We know the current I will have this
same form: I пЂЅ ImпѓќCe iпЃ§t пѓћ. The relations between the voltage and the current are linear, and
so we can consider complex voltages and currents and use the fact that
e iпЃ§t пЂЅ cos пЃ§t пЂ« i sin пЃ§t. We thus assume a more or less fictional complex voltage V , the
imaginary part of which is the actual voltage, and then the actual current will be the
imaginary part of the resulting complex current.

What makes this a good idea is the fact that differentiation with respect to time t becomes
d
simply multiplication by iпЃ§: dt Ae iпЃ§t пЂЅ iпЃ§Ae iпЃ§t . If I пЂЅ be iпЃ§t , the above relations between
I
current and voltage become V пЂЅ iпЃ§LI for an inductor, and iпЃ§VC пЂЅ I, or V пЂЅ iпЃ§C for a
capacitor. Calculus is thereby turned into algebra. To illustrate, suppose we have a simple
RLC circuit with a voltage source V пЂЅ a sin пЃ§t. We let E пЂЅ ae iwt .

Then the fact that the voltage drop around a closed circuit must be zero (one of KirchoffвЂ™s
celebrated laws) looks like

3.2
I пЂ« RI пЂЅ ae iпЃ§t , or
iпЃ§LI пЂ«
iпЃ§C
iпЃ§Lb пЂ« b пЂ« Rb пЂЅ a
iпЃ§C

Thus,
a
bпЂЅ .
1
R пЂ« i пЃ§L пЂї пЃ§C

In polar form,

a e iпЃЄ ,
bпЂЅ
2
1
R 2 пЂ« пЃ§L пЂї пЃ§C

where

1
пЃ§L пЂї пЃ§C
tan пЃЄ пЂЅ . (R п‚® 0)
R

Hence,

a
I пЂЅ Imпѓќbe iпЃ§t пѓћ пЂЅ Im e iпѓќпЃ§tпЂ«пЃЄпѓћ
2
1
R 2 пЂ« пЃ§L пЂї пЃ§C
a
пЂЅ sinпѓќпЃ§t пЂ« пЃЄпѓћ
2
1
R 2 пЂ« пЃ§L пЂї пЃ§C

This result is well-known to all, but I hope you are convinced that this algebraic approach
afforded us by the use of complex numbers is far easier than solving the differential
equation. You should note that this method yields the steady state solutionвЂ”the transient
solution is not necessarily sinusoidal.

Exercises

1. Show that expпѓќz пЂ« 2пЃћiпѓћ пЂЅ expпѓќzпѓћ.

expпѓќzпѓћ
пЂЅ expпѓќz пЂї wпѓћ.
2. Show that expпѓќwпѓћ

3. Show that |expпѓќzпѓћ| пЂЅ e x , and argпѓќexpпѓќzпѓћпѓћ пЂЅ y пЂ« 2kпЃћ for any argпѓќexpпѓќzпѓћпѓћ and some

3.3
integer k.

4. Find all z such that expпѓќzпѓћ пЂЅ пЂї1, or explain why there are none.

5. Find all z such that expпѓќzпѓћ пЂЅ 1 пЂ« i, or explain why there are none.

6. For what complex numbers w does the equation expпѓќzпѓћ пЂЅ w have solutions? Explain.

7. Find the indicated mesh currents in the network:

3.3 Trigonometric functions. Define the functions cosine and sine as follows:

пЂїiz
iz
cos z пЂЅ e пЂ« e ,
2
пЂїiz
iz
sin z пЂЅ e пЂї e
2i

where we are using e z пЂЅ expпѓќzпѓћ.

First, letвЂ™s verify that these are honest-to-goodness extensions of the familiar real functions,
cosine and sineвЂ“otherwise we have chosen very bad names for these complex functions.
So, suppose z пЂЅ x пЂ« 0i пЂЅ x. Then,

e ix пЂЅ cos x пЂ« i sin x, and
e пЂїix пЂЅ cos x пЂї i sin x.

Thus,

3.4
пЂїix
ix
cos x пЂЅ e пЂ« e ,
2
пЂїix
ix
sin x пЂЅ e пЂї e ,
2i

and everything is just fine.

Next, observe that the sine and cosine functions are entireвЂ“they are simply linear
combinations of the entire functions e iz and e пЂїiz . Moreover, we see that

d sin z пЂЅ cos z, and d пЂЅ пЂї sin z,
dz dz

just as we would hope.

It may not have been clear to you back in elementary calculus what the so-called
hyperbolic sine and cosine functions had to do with the ordinary sine and cosine functions.
Now perhaps it will be evident. Recall that for real t,

пЂїt пЂїt
t t
sinh t пЂЅ e пЂї e , and cosh t пЂЅ e пЂ« e .
2 2

Thus,
пЂїiпѓќitпѓћ пЂїt
iпѓќitпѓћ t
sinпѓќitпѓћ пЂЅ e пЂї e пЂЅ i e пЂї e пЂЅ i sinh t.
2
2i

Similarly,
cosпѓќitпѓћ пЂЅ cosh t.

How nice!

Most of the identities you learned in the 3 rd grade for the real sine and cosine functions are
also valid in the general complex case. LetвЂ™s look at some.

sin 2 z пЂ« cos 2 z пЂЅ 1 пѓџпЂїпѓќe iz пЂї e пЂїiz пѓћ 2 пЂ« пѓќe iz пЂ« e пЂїiz пѓћ 2 пѓ
4
пЂЅ 1 пѓџпЂїe 2iz пЂ« 2e iz e пЂїiz пЂї e пЂї2iz пЂ« e 2iz пЂ« 2e iz e пЂїiz пЂ« e пЂї2iz пѓ
4
пЂЅ 1 пѓќ2 пЂ« 2пѓћ пЂЅ 1
4

3.5
It is also relative straight-forward and easy to show that:

sinпѓќz п‚± wпѓћ пЂЅ sin z cos w п‚± cos z sin w, and
cosпѓќz п‚± wпѓћ пЂЅ cos z cos w пЃЂ sin z sin w

Other familiar ones follow from these in the usual elementary school trigonometry fashion.

LetвЂ™s find the real and imaginary parts of these functions:

sin z пЂЅ sinпѓќx пЂ« iyпѓћ пЂЅ sin x cosпѓќiyпѓћ пЂ« cos x sinпѓќiyпѓћ
пЂЅ sin x cosh y пЂ« i cos x sinh y.

In the same way, we get cos z пЂЅ cos x cosh y пЂї i sin x sinh y.

Exercises

8. Show that for all z,
пЃћ
a)sinпѓќz пЂ« 2пЃћпѓћ пЂЅ sin z; b)cosпѓќz пЂ« 2пЃћпѓћ пЂЅ cos z; c)sin z пЂ« пЂЅ cos z.
2

9. Show that |sin z| 2 пЂЅ sin 2 x пЂ« sinh 2 y and |cos z| 2 пЂЅ cos 2 x пЂ« sinh 2 y.

10. Find all z such that sin z пЂЅ 0.

11. Find all z such that cos z пЂЅ 2, or explain why there are none.

3.4. Logarithms and complex exponents. In the case of real functions, the logarithm
function was simply the inverse of the exponential function. Life is more complicated in
the complex caseвЂ”as we have seen, the complex exponential function is not invertible.
There are many solutions to the equation e z пЂЅ w.

If z п‚® 0, we define log z by

log z пЂЅ ln|z| пЂ« i arg z.

There are thus many log zвЂ™s; one for each argument of z. The difference between any two of
these is thus an integral multiple of 2пЃћi. First, for any value of log z we have

3.6
e log z пЂЅ e ln |z|пЂ«i arg z пЂЅ e ln |z| e i arg z пЂЅ z.

This is familiar. But next there is a slight complication:

logпѓќe z пѓћ пЂЅ ln e x пЂ« i arg e z пЂЅ x пЂ« пѓќy пЂ« 2kпЃћ пѓћi
пЂЅ z пЂ« 2kпЃћi,

where k is an integer. We also have

logпѓќzwпѓћ пЂЅ lnпѓќ|z||w|пѓћ пЂ« i argпѓќzwпѓћ
пЂЅ ln |z| пЂ« i arg z пЂ« ln |w| пЂ« i arg w пЂ« 2kпЃћi
пЂЅ log z пЂ« log w пЂ« 2kпЃћi

for some integer k.

There is defined a function, called the principal logarithm, or principal branch of the
logarithm, function, given by

Log z пЂЅ ln|z| пЂ« iArg z,

where Arg z is the principal argument of z. Observe that for any log z, it is true that
log z пЂЅLog z пЂ« 2kпЃћi for some integer k which depends on z. This new function is an
extension of the real logarithm function:

Log x пЂЅ ln x пЂ« iArg x пЂЅ ln x.

This function is analytic at a lot of places. First, note that it is not defined at z пЂЅ 0, and is
not continuous anywhere on the negative real axis (z пЂЅ x пЂ« 0i, where x пЂј 0.). So, letвЂ™s
suppose z 0 пЂЅ x 0 пЂ« iy 0 , where z 0 is not zero or on the negative real axis, and see about a
derivative of Log z :

Log z пЂї Log z 0 Log z пЂї Log z 0
пЂЅlim
lim .
z пЂї z0 zп‚ёz 0 e Log z пЂї e Log z 0
zп‚ёz 0

Now if we let w пЂЅLog z and w 0 пЂЅLog z 0 , and notice that w п‚ё w 0 as z п‚ё z 0 , this becomes

3.7
Log z пЂї Log z 0
пЂЅ lim w пЂї ww 00
lim z пЂї z0 w
wп‚ёw 0 e пЂї e
zп‚ёz 0

пЂЅ 1 0 пЂЅ z0 1
w
e

1
Thus, Log is differentiable at z 0 , and its derivative is .
z0

We are now ready to give meaning to z c , where c is a complex number. We do the obvious
and define
z c пЂЅ e c log z .

There are many values of log z, and so there can be many values of z c . As one might guess,
e cLog z is called the principal value of z c .

Note that we are faced with two different definitions of z c in case c is an integer. LetвЂ™s see
if we have anything to unlearn. Suppose c is simply an integer, c пЂЅ n. Then

z n пЂЅ e n log z пЂЅ e kпѓќLog zпЂ«2kпЃћi пѓћ
пЂЅ e nLog z e 2knпЃћi пЂЅ e nLog z

There is thus just one value of z n , and it is exactly what it should be: e nLog z пЂЅ |z| n e in arg z . It
is easy to verify that in case c is a rational number, z c is also exactly what it should be.

Far more serious is the fact that we are faced with conflicting definitions of z c in case
z пЂЅ e. In the above discussion, we have assumed that e z stands for expпѓќzпѓћ. Now we have a
definition for e z that implies that e z can have many values. For instance, if someone runs at
you in the night and hands you a note with e 1/2 written on it, how to you know whether this
means expпѓќ1/2пѓћ or the two values e and пЂї e ? Strictly speaking, you do not know. This
ambiguity could be avoided, of course, by always using the notation expпѓќzпѓћ for e x e iy , but
almost everybody in the world uses e z with the understanding that this is expпѓќzпѓћ, or
equivalently, the principal value of e z . This will be our practice.

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