5 Stereographic Projection; The Point at Infinity

1.5 Stereographic Projection; The Point at Infinity



17



that is, let be the sphere in Euclidean (ξ, η, ζ ) space with distance 12 from (0, 0, 12 ).

Suppose, moreover, that the plane ζ = 0 coincides with the complex plane C, and that

the ξ and η axes are the x and y axes, respectively. To each (ξ, η, ζ ) ∈ we associate

the complex number z where the ray from (0, 0, 1) through (ξ, η, ζ ) intersects C.

This establishes a 1-1 correspondence, known as stereographic projection, between

C and the points of

other than (0, 0, 1). Formulas governing this correspondence

can be derived as follows. Since (0, 0, 1), (ξ, η, ζ ) and (x, y, 0) are collinear,

x

y

1

= =

ξ

η

1−ζ

ζ

(0, 0, 1)



(ξ, η, ζ )



η=y



(0, 0, 0)



(x, y, 0)



ξ =x



so that

x=



ξ

;

1−ζ



y=



η

.

1−ζ



(2)



We leave it as an exercise to show that the equations (1) and (2) can be solved for

ξ, η, ζ in terms of x, y as

ξ=



x

;

x 2 + y2 + 1



η=



y

;

x 2 + y2 + 1



ζ =



x 2 + y2

.

x 2 + y2 + 1



(3)



Now suppose that {σk } = {(ξk , ηk , ζk )} is a sequence of points of which converges

to (0, 0, 1) and let {z k } be the corresponding sequence in C. By (2),

x 2 + y2 =



ξ 2 + η2

ζ

,

=

1−ζ

(1 − ζ )2



so that as σk → (0, 0, 1), |z k | → ∞. Conversely, it follows from (3) that if |z k| → ∞,

σk → (0, 0, 1). Loosely speaking, this suggests that the point (0, 0, 1) on

corresponds to ∞ in the complex plane. We can make this more precise by formally

adjoining to C a “point at infinity” and defining its neighborhoods as the sets in

C corresponding to the spherical neighborhoods of (0, 0, 1). (See Exercise 24.)



18



1 The Complex Numbers



While we will not examine the resulting “extended plane” in greater detail, we will

adopt the following convention.

1.11 Definition

We say {z k } → ∞ if |z k | → ∞; i.e., |z k | → ∞ if for any M > 0, there exists

an integer N such that k > N implies |z k | > M. Similarly, we say f (z) → ∞ if

| f (z)| → ∞.

For future reference, we note the connection between circles on

and circles

in C. By a circle on , we mean the intersection of

with a plane of the form

Aξ + Bη +Cζ = D. According to (3), if S is such a circle and T is the corresponding

set in C,

(C − D)(x 2 + y 2 ) + Ax + By = D

(4)

for (x, y) ∈ T . Note that if C = D, (4) is the equation of a circle. If C = D, (4)

represents a line. Since C = D if and only if S intersects (0, 0, 1), we have the

following proposition.

1.12 Proposition

Let S be a circle on



and let T be its projection on C. Then



a. if S contains (0, 0, 1), T is a line;

b. if S doesn’t contain (0, 0, 1), T is a circle.

The converse of Proposition 1.12 is also valid. We leave its proof as an exercise.

(See Exercise 25.)



Exercises

1. Express in the form a + bi:

1

a.

6 + 2i

√ 4

1

3

c. − + i

2

2



b.



(2 + i)(3 + 2i)

1−i



d. i 2 , i 3 , i 4 , i 5 , . . .



2. Find (in rectangular form) the two values of

√

3. Solve the equation z 2 + 32 iz − 6 i = 0.



√

−8 + 6 i.



4. Prove the following identities:

a. z 1 + z 2 = z l + z 2 .

b. z 1 z 2 = z 1 · z 2 .

c. P(z) = P(¯z ), for any polynomial P with real coefficients.

d. z¯¯ = z.

5. Suppose P is a polynomial with real coefficients. Show that P(z) = 0 if and only if P(¯z ) = 0 [i.e.,

zeroes of “real” polynomials come in conjugate pairs].

6. Verify that |z 2 | = |z|2 using rectangular coordinates and then using polar coordinates.



Exercises



19



7. Show

a. |z n | = |z|n .

b. |z|2 = z z¯ .

c. |Re z|, |Im z| ≤ |z| ≤ |Re z| + |Im z|.

(When is equality possible?)

8. a. Fill in the details of the following proof of the triangle inequality:

|z 1 + z 2 |2 = (z 1 + z 2 )(z 1 + z 2 )

= |z 1 |2 + |z 2 |2 + z 1 z 2 + z 1 z 2

= |z 1 |2 + |z 2 |2 + 2 Re(z 1 z 2 )

≤ |z 1 |2 + |z 2 |2 + 2|z 1 ||z 2 |

= (|z 1 | + |z 2 |)2 .

b. When can equality occur?

c. Show: |z 1 − z 2 | ≥ |z 1 | − |z 2 |.

9.* It is an interesting fact that a product of two sums of squares is itself a sum of squares. For example,

(12 + 22 )(32 + 42 ) = 125 = 52 + 102 = 22 + 112 .

a. Prove the result using complex algebra. That is, show that for any two pairs of integers {a, b} and

{c, d}, we can find integers u, v with

(a 2 + b2 )(c2 + d 2 ) = u 2 + v 2

b. Show that, if a, b, c, d are all nonzero and at least one of the sets {a 2 , b2 } and {c2 , d 2 } consists

of distinct positive integers, then we can find u 2 , v 2 as above with u 2 and v 2 both positive.

c. Show that, if a, b, c, d are all nonzero and both of the sets {a 2 , b2 } and {c2 , d 2 } consist of distinct

positive integers, then there are two different sets {u 2 , v 2 } and {s 2 , t 2 } with

(a 2 + b2 )(c2 + d 2 ) = u 2 + v 2 = s 2 + t 2 .

d. Give a geometric interpretation and proof of the results in b) and c), above.

10.* Prove: |z 1 + z 2 |2 + |z 1 − z 2 |2 = 2(|z 1 |2 + |z 2 |2 ) and interpret the result geometrically.

11. Let z = x + iy. Explain the connection between Arg z and tan−1 (y/x). (Warning: they are not

identical.)

12. Solve the following equations in polar form and locate the roots in the complex plane:

a. z 6 = 1.

b. z 4 = −1. √

c. z 4 = −1 + 3i.

13. Show that the n-th roots of 1 (aside from 1) satisfy the “cyclotomic” equation z n−1 + z n−2 + · · · +

z + 1 = 0. [Hint: Use the identity z n − 1 = (z − 1)(z n−1 + z n−2 + · · · + 1).]

14. Suppose we consider the n − 1 diagonals of a regular n-gon inscribed in a unit circle obtained by

connecting one vertex with all the others. Show that the product of their lengths is n. [Hint: Let the

vertices all be connected to 1 and apply the previous exercise.]



20



1 The Complex Numbers



15. Describe the sets whose points satisfy the following relations. Which of the sets are regions?

z−1

= 1.

a. |z − i| ≤ 1.

b.

z+1

c. |z − 2| > |z − 3|.

e.



d. |z| < 1 and Im z > 0.



1

= z¯ .

z



f. |z|2 = Im z.



g. |z 2 − 1| < 1. [Hint: Use polar coordinates.]

16.* Identify the set of points which satisfy

a. |z| = Rez + 1

b. |z − 1| + |z + 1| = 4



c. z n−1 = z



17. Let Arg w denote that value of the argument between −π and π (inclusive). Show that

Arg



z−1

z+1



=



π/2



if Im z > 0



−π/2 if Im z < 0



where z is a point on the unit circle |z| = 1.

√

18.* √

Find the three roots of x 3 − 6x = 4 by finding the three real-valued possibilities for 3 2 + 2i +

3

2 − 2i.(Note: You can find the three cube roots of 2 + 2i, or you can simplify the problem by first

applying the identity: a + b = (a 3 + b3 )/(a 2 − ab + b2 ).

19.* Prove that x 3 + px = q has three real roots if and only if 4 p3 < −27q 2 . (Hint: Find the local

minimum and local maximum values of x 3 + px − q.)

20.* a. Let P(z) = 1 + 2z + 3z 2 + · · · + nz n−1 . By considering (1 − z)P(z), show that all the zeroes

of P(z) are inside the unit disc.

b. Show that the same conclusion applies to any polynomial of the form: a0 +a1 z+a2 z 2 +···+an z n ,

with all ai real and 0 ≤ a0 ≤ a1 ≤ · · · ≤ an

21. Show that

a. f (z) =



∞ kz k is continuous in |z| < 1.

k=0



b. g(z) =



∞ 1/(k 2 + z) is continuous in the right half-plane Re z > 0.

k=1



22. Prove that a polygonally connected set is connected.

23. Let

S = x + iy : x = 0 or x > 0, y = sin



1

.

x



Show that S is connected, even though there are points in S that cannot be connected by any curve

in S.

24. Let S = {(ξ, η, ζ ) ∈

: ζ ≥ ζ0 }, where 0 < ζ0 < 1 and let T be the corresponding set in C. Show

that T is the exterior of a circle centered at 0.

25. Suppose T ⊂ C. Show that the corresponding set S ⊂

a. a circle if T is a circle.

b. a circle minus (0, 0, 1) if T is a line.



is



26. Let P be a nonconstant polynomial in z. Show that P(z) → ∞ as z → ∞.

27. Suppose that z is the stereographic projection of (ξ, η, ζ ) and 1/z is the projection of (ξ , η , ζ ).

a. Show that (ξ , η , ζ ) = (ξ, −η, 1 − ζ ).

b. Show that the function 1/z, z ∈ C, is represented on

by a 180◦ rotation about the diameter

with endpoints (− 21 , 0, 12 ) and ( 12 , 0, 12 ).

28. Use exercise (27) to show that f (z) = 1/z maps circles and lines in C onto other circles and lines.



Chapter 2



Functions of the Complex Variable z



Introduction

We wish to examine the notion of a “function of z” where z is a complex variable. To

be sure, a complex variable can be viewed as nothing but a pair of real variables so

that in one sense a function of z is nothing but a function of two real variables. This

was the point of view we took in the last section in discussing continuous functions.

But somehow this point of view is too general. There are some functions which are

“direct” functions of z = x + i y and not simply functions of the separate pieces x

and y.

Consider, for example, the function x 2 − y 2 + 2i x y. This is a direct function of

x + i y since x 2 − y 2 + 2i x y = (x + i y)2 ; it is the function squaring. On the other

hand, the only slightly different-looking function x 2 + y 2 − 2i x y is not expressible

as a polynomial in x + i y. Thus we are led to distinguish a special class of functions,

those given by direct or explicit or analytic expressions in x + i y. When we finally

do evolve a rigorous definition, these functions will be called the analytic functions.

For now we restrict our attention to polynomials.



2.1 Analytic Polynomials

2.1 Definition

A polynomial P(x, y) will be called an analytic polynomial if there exist (complex)

constants αk such that

P(x, y) = α0 + α1 (x + i y) + α2 (x + i y)2 + · · · + α N (x + i y) N .

We will then say that P is a polynomial in z and write it as

P(z) = α0 + α1 z + α2 z 2 + · · · + α N z N .



21



22



2 Functions of the Complex Variable z



x2



Indeed, x 2 − y 2 + 2i x y is analytic. On the other hand, as we mentioned above,

+ y 2 − 2i x y is not analytic, and we now prove this assertion. So suppose

N



x + y − 2i x y ≡

2



αk (x + i y)k .



2



k=0



Setting y = 0, we obtain

N



x2 ≡



αk x k

k=0



or

α0 + α1 x + (α2 − 1)x 2 + · · · + α N x N ≡ 0.

Setting x = 0 gives α0 = 0; dividing out by x and again setting x = 0 shows α1 = 0,

etc. We conclude that

α1 = α3 = α4 = · · · = α N = 0

α2 = 1,

and so our assumption that

N



x 2 + y 2 − 2i x y ≡



αk (x + i y)k

k=0



has led us to

x 2 + y 2 − 2i x y ≡ (x + i y)2 = x 2 − y 2 + 2i x y,

which is simply false!

A bit of experimentation, using the method described above (setting y = 0 and

“comparing coefficients”) will show how rare the analytic polynomials are. A randomly chosen polynomial, P(x, y), will hardly ever be analytic.

E XAMPLE

x 2 + i v(x, y) is not analytic for any choice of the real polynomial v(x, y). For

a polynomial in z can have a real part of degree 2 in x only if it is of the form

az 2 + bz + c with a = 0. In that case, however, the real part must contain a y 2 term

as well.

♦

Another Way of Recognizing Analytic Polynomials We have seen, in our method of

comparing coefficients, a perfectly adequate way of determining whether a given

polynomial is or is not analytic. This method, we point out, can be condensed to the

statement: P(x, y) is analytic if and only if P(x, y) = P(x +i y, 0). Looking ahead to



2.1 Analytic Polynomials



23



the time we will try to extend the notion of “analytic” beyond the class of polynomials,

however, we see that we can expect trouble! What is so simple for polynomials is

totally intractable for more general functions. We can evaluate P(x +i y, 0) by simple

arithmetic operations, but what does it mean to speak of f (x + i y, 0)? For example,

if f (x, y) = cos x + i sin y, we observe that f (x, 0) = cos x. But what shall we

mean by cos(x + i y)? What is needed is another means of recognizing the analytic

polynomials, and for this we retreat to a familiar, real-variable situation. Suppose

that we ask of apolynomial P(x, y) whether it is a function of the single variable

x + 2y. Again the answer can be given in the spirit of our previous one, namely:

P(x, y) is a function of x + 2y if and only if P(x, y) = P(x + 2y, 0). But it can also

be given in terms of partial derivatives! A function of x + 2y undergoes the same

change when x changes by ∈ as when y changes /2 and this means exactly that its

partial derivative with the respect to y is twice its partial derivative with respect to x.

That is, P(x, y) is a function of x + 2y if and only if Py = 2Px .

Of course, the “2” can be replaced by any real number, and we obtain the more

general statement: P(x, y) is a function of x + λy if and only if Py = λPx .

Indeed for polynomials, we can even ignore the limitation that λ be real, which

yields the following proposition.

2.2 Definition

Let f (x, y) = u(x, y)+i v(x, y) where u and v are real-valued functions. The partial

derivatives f x and f y are defined by u x + i v x and u y + i v y respectively, provided

the latter exist.

2.3 Proposition

A polynomial P(x, y) is analytic if and only if Py = i Px .

Proof

The necessity of the condition can be proven in a straightforward manner. We leave

the details as an exercise. To show that it is also sufficient, note that if

Py = i Px ,

the condition must be met separately by the terms of any fixed degree. Suppose then

that P has n-th degree terms of the form

Q(x, y) = C0 x n + C1 x n−1 y + C2 x n−2 y 2 + · · · + Cn y n .

Since

Q y = i Qx ,

C1 x n−1 + 2C2 x n−2 y + · · · + nCn y n−1

= i [nC0 x n−1 + (n − 1)C1 x n−2 y + · · · + Cn−1 y n−1 ].



24



2 Functions of the Complex Variable z



Comparing coefficients,

C1 = i nC0 = i

C2 =



n

1



C0



i (n − 1)

n(n − 1)

C1 = i 2

C0 = i 2

2

2



n

2



C0 ,



and in general

Ck = i k



n

k



C0



so that

n



n



n

k



Ck x n−k y k = C0



Q(x, y) =

k=0



k=0



x n−k (i y)k = C0 (x + i y)n .



Thus P is analytic.

The condition f y = i f x is sometimes given in terms of the real and imaginary

parts of f . That is, if f = u + i v, then

fx = u x + i vx

fy = uy + ivy

and the equation f y = i f x is equivalent to the twin equations

ux = vy;



u y = −v x .



(1)



These are usually called the Cauchy-Riemann equations.

E XAMPLES

1. A non-constant analytic polynomial cannot be real-valued, for then both Px and

Py would be real and the Cauchy-Riemann equations would not be satisfied.

2. Using the Cauchy-Riemann equations, one can verify that x 2 − y 2 + 2i x y is

analytic while x 2 + y 2 − 2i x y is not.

♦

Finally, we note that polynomials in z have another property which distinguishes

them as functions of z: they can be differentiated directly with respect to z. We will

make this more precise below.

2.4 Definition

A complex-valued function f , defined in a neighborhood of z, is said to be differentiable at z if

lim



h→0



f (z + h) − f (z)

h



exists. In that case, the limit is denoted f (z).



2.2 Power Series



25



It is important to note that in Definition 2.4, h is not necessarily real. Hence the

limit must exist irrespective of the manner in which h approaches 0 in the complex

plane. For example, f (z) = z¯ is not differentiable at any point z since

f (z + h) − f (z)

h¯

=

h

h

which equals +1 if h is real and −1 if h is purely imaginary.

2.5 Proposition

If f and g are both differentiable at z, then so are

h1 = f + g

h2 = f g

and, if g(z) = 0,

h3 =



f

.

g



In the respective cases,

h 1 (z) = f (z) + g (z)

h 2 (z) = f (z)g(z) + f (z)g (z)

h 3 (z) = [ f (z)g(z) − f (z)g (z)]/g 2 (z).



Proof

Exercise 6.

2.6 Proposition

If P(z) = α0 + α1 z + · · · + α N z N , then P is differentiable at all points z and

P (z) = α1 + 2α2 z + · · · + Nα N z N−1 .

Proof

See Exercise 7.



2.2 Power Series

We now consider a wider class of direct functions of z–those given by infinite polynomials or “power series” in z.



26



2 Functions of the Complex Variable z



2.7 Definition

k

A power series in z is an infinite series of the form ∞

k=0 Ck z .

To study the convergence of a power series, we recall the notion of the lim of a

positive real-valued sequence. That is,



lim an = lim



n→∞



n→∞



sup ak .

k≥n



Since supk≥n ak is a non-increasing function of n, the limit always exists or equals

+∞. The properties of the lim which will be of interest to us are the following.

If limn→∞ an = L,

i. for each N and for each > 0, there exists some k > N such that ak ≥ L − ;

ii. for each > 0, there is some N such that ak ≤ L + for all k > N.

iii. lim can = cL for any nonnegative constant c.

2.8 Theorem

Suppose lim|Ck |1/k = L.

1. If L = 0, Ck z k converges for all z.

2. If L = ∞, Ck z k converges for z = 0 only.

3. If 0 < L < ∞, set R = 1/L. Then Ck z k converges for |z| < R and diverges

for |z| > R. (R is called the radius of convergence of the power series.)

Proof

1. L = 0.

Since lim|Ck |1/k = 0, lim|Ck |1/k |z| = 0 for all z. Thus, for each z, there is some

N such that k > N implies

|Ck z k | ≤



1

,

2k



so that |Ck z k | converges; therefore, by the Absolute Convergence Test,

converges.

2. L = ∞.

For any z = 0,

|Ck |1/k ≥



Ck z k



1

|z|



for infinitely many values of k. Hence |Ck z k | ≥ 1, the terms of the series do

not approach zero, and the series diverges. (The fact that the series converges for

z = 0 is obvious.)



2.2 Power Series



27



3. 0 < L < ∞, R = 1/L.

Assume first that |z| < R and set |z| = R(1 − 2δ). Then since lim|Ck |1/ k |z| =

(1 − 2δ), |Ck |1/k |z| < 1 − δ for sufficiently large k and

Ck z k is absolutely

1/k

convergent. On the order hand, if |z| > R, lim|Ck | |z| > 1, so that for infinitely many values of k, Ck z k has absolute value greater than 1 and

Ck z k

diverges.

k

Note that if ∞

k=0 Ck z has radius of convergence R, the series converges uniformly in any smaller disc: |z| ≤ R − δ. For then

∞



∞



|Ck z k | ≤

k=0



|Ck |(R − δ)k ,

k=0



which also converges. Hence a power series is continuous throughout its domain of

convergence. (See Theorem 1.9.)

All three cases above can be combined by noting that a power series always

converges inside a disc of radius

R = 1/lim|Ck |1/k .

Here R = 0 means that the series converges at z = 0 only and R = ∞ means that

the series converges for all z. In the cases where 0 < R < ∞, while the theorem

assures us that the series diverges for |z| > R, it says nothing about the behavior of

the power series on the circle of convergence |z| = R. As the following examples

demonstrate, the series may converge for all or some or none of the points on the

circle of convergence.

E XAMPLES

n

1. Since n 1/n → 1, ∞

n=1 nz converges for |z| < 1 and diverges for |z| > 1. The

series also diverges for |z| = 1 for then |nz n | = n → ∞. (See Exercise 8.)

∞

n

2

2.

n=1 (z /n ) also has radius of convergence equal to 1. In this case, however,

the series converges for all points z on the unit circle since

zn

1

= 2

n2

n

3.

4.



for |z| = 1.



∞

n

n=1 (z /n) has radius of convergence equal to 1. In this case, the series converges



at all points of the unit circle except z = 1. (See Exercise 12.)

∞

n

n=0 (z /n!) converges for all z since

1

→ 0.

(n!)1/n



5.



(See Exercise 13.)

∞

n n n

n=0 [1 + (−1) ] z has radius of convergence

lim 2 = 2.



1

2



since lim[1 + (−1)n ] =