INTRODUCTION TO COMPLEX NUMBERS INTRODUCTION TO MAPLE

INTRODUCTION TO MAPLE

September 2002

Faculty ITS

Delft University of Technology

Complex Numbers

The equation x² − 1 = 0 has two solutions: x = 1 and x = −1. Since 1 and −1 are real numbers, we call these solutions real solutions. The equation x²+ 1 = 0 has no real solutions: there is no real number x such that x²+ 1 = 0. To fix this problem we extend the set^R of real numbers with a formal solution of the equation x²+ 1 = 0, which we will denote by the symbol i. With this symbol we will construct a set of new numbers.

A complex number is an expression of the form a + bi, where a and b are real numbers.

If a + bi is a complex number, then a is called the real part and b the imaginary part of this number. For instance, the real part of the complex number−3 + 4i is equal to −3 and the imaginary part is equal to 4 (which is real!!).

The set of all complex numbers is written as^C. So we may abbreviate ‘a + bi is a complex number’ to ‘a + bi ∈^C’.

Note that both the real and the imaginary part of a complex number are real numbers.

Two complex numbers a + bi and c + di are equal if the real and imaginary part are equal:

a + bi = c + di ⇐⇒ a = c en b = d.

Each real number a can be considered as a complex number with zero imaginary part:

a = a + 0i.

So the real numbers form a subset of the complex numbers. In particular we will write 0 for 0 + 0i.

A complex number with zero real part will be called purely imaginary .

Re-axis Im-axis

O i 3+i

−2+2i 1+2i

3−i

Figure 0.1: The Complex Plane

The set^R of real numbers can be represented graphically by a straight line: the real line.

Each point on this line corresponds with a real number.

The set ^C of complex numbers can also be represented graphically, by a plane, called the complex plane or Argand plane. The point (a, b) in this plane corresponds with the complex number a + bi. The horizontal axis consists of all complex numbers of the form a + 0i, and corresponds therefore with the real line; therefore this axis is called the real axis. Likewise the vertical axis corresponding with the complex numbers of the form 0 + bi is called the imaginary axis.

The sum of two complex numbers is defined by adding their real parts and their imaginary parts: (a + bi) + (c + di) = (a + c) + (b + d)i. For the product we use that

i² = i · i = −1

since the symbol i is a formal solution of x² = −1. Thanks to this rule two complex numbers can be multiplied by “eliminating parentheses” : (a + bi) · (c + di) = ac + adi + bci + bdi² = (ac − bd) + (ad + bc)i. Summarizing:

(a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) · (c + di) = (ac − bd) + (ad + bc)i

Re-axis Im-axis

O

3+i

1+2i 4+3i=(3+i)+(1+2i)

3−i=3+i

Figure 0.2: The Complex Plane: Addition and Complex Conjugation

Plotting complex numbers a + bi as vectors (a, b) in the complex plane, the addition of two complex numbers a + bi and c + di comes down to vector addition of (a, b) and (c, d).

We define for a complex number z = a + bi its complex conjugate to be ¯z = a − bi. In the complex plane taking conjugates corresponds to reflecting about the real axis.

Example 0.1. Fractions can be brought conveniently into the form a + bi by multiplying numerator and denominator by the complex conjugate of the denominator:

1− i

2 + i = 1− i 2 + i·2− i

2− i = (1− i)(2 − i)

(2 + i)(2 − i) = 2− 3i + i²

4− i² = 1− 3i

5 = ¹₅ −³₅i.

Conjugating satisfies the following properties.

Theorem 0.2 (Rules for Complex Conjugation). For all z ∈^C and w ∈^C: z + w = z + w; zw = z w; zⁿ= zⁿ; z = z.

Proof. See Stewart App.G exercise 18.

The modulus or absolute value |z| of a complex number z = a + bi is its distance in the complex plane from the origin. By the Pythagoras theorem we see:

|z| =
a²+ b²

b z = a + ib

a Im-axis

Re-axis u-v i

u u-v v

z

Figure 0.3: The modulus |z| of the complex number z.

If u and v are two complex numbers, we can interpret the modulus of their difference,

|u − v|, geometrically as the distance from the point u to the point v in the complex plane (cf. figure 0.3).

Notice that (a + bi) · (a − bi) = a²− b²i²= a²+ b² and so z¯z = |z|²

In particular this explains why the division procedure of example 0.1 works in general: if w = 0, then |w|² is a positive real number, so

z w = zw

ww = 1

|w|² zw can be written directly in the form a + bi.

Finally we make a remark about the notation. Sometimes we will write i =√

−1, since i² = −1. But notice that we also have (−i)² = i² = −1 and so −i is also a square root of −1. We say that i is the principal square root of −1. In general, if c is a positive

number, we write √

−c =√ c i.

However: the well-known rules for taking roots do not hold anymore for roots of negative numbers! As an illustration we “derive” the equality −1 = 1:

−1 = i² = i · i =√

−1 ·√

−1 =√

−1 · −1 =√ 1 = 1.¹

Except for this anomaly we can do arithmetic with complex numbers just as with real numbers, each time using i² = −1. Like with real numbers we define for a complex number z = 0 the following notation: z⁰ = 1 and z^−k= _z¹_k (for an integer k ≥ 1).

POLAR COORDINATES In a rectangular coordinate system with x-axis and y- axis is the position of a point P = (a, b) fully determined by the x-coordinate a and the y-coordinate b. The pair (a, b) is called the Cartesian² coordinates of the point P . Instead of Cartesian coordinates one uses often polar coordinates: P = (r, ϕ). In polar

1Of course the 4th =-sign is false.

2Cartesius is the Latin name of R. Descartes (1586-1650)

coordinates is r the distance from P to the origin O and ϕ the angle (measured in the counterclockwise direction) between the positive x-axis and the line OP (cf. figure 0.4).

W e have r =√

a²+ b²; in particular³ r ≥ 0. The angle ϕ is not unique. For points P = O is ϕ fixed up to a multiple of 2π and for P = O we have no angle ϕ.

x-as y-as

O

ϕ b

a r

P =(a,b)

Figure 0.4: The polar coordinates of the point P = (a, b) W e call r the polar radius of P and ϕ the polar angle of P .

Between Cartesisian and polar coordinates we have the following connection. We get the Cartesisian coordinates (a, b) from a point P with polar coordinates (r, ϕ) by the compution rules

a = r cos ϕ b = r sin ϕ

Given the Cartesian coordinates (a, b) from a point P not equal to O we can get the polar coordinates (r, ϕ) from

r =
a²+ b² cos ϕ = √ a

a²+ b² sin ϕ = √ b

a²+ b²

Note that indeed ϕ is fixed by these formulas only up to a multiple of 2π. For each P not equal to O there is exactly one polar angle ϕ satisfying −π < ϕ ≤ π. Further we have tan ϕ = _a^b (but this formula does not fix ϕ up to a multiple of 2π!).

The tangent function repeates itself after a period π, so for each x we can find an angle ϕ such that tan ϕ = x, with −^π₂ < ϕ < ^π₂. We call this angle arctan x, the arctangent of x.

So arctan 0 = 0, arctan 1 = π/4 and arctan(−1) = −π/4. With this new function we have tan ϕ = x ⇐⇒ ϕ = arctan x + k · π, with k in ^Z.

It is often convenient to express the polar angle of a point as an arctangent. If (a, b) = (r cos ϕ, r sin ϕ) with −π < ϕ ≤ π and if a = 0, then tan ϕ = _a^b, so either ϕ = arctan^b_a or ϕ = π + arctan^b_a or ϕ = −π + arctan_a^b. Now there are no other possibilities, because of the limits for ϕ and the arctangent. Which of these three possibilities is the case depends on the quadrant in which (a, b) lies.

Example 0.3. We choose again the polar angle ϕ such that −π < ϕ ≤ π. What is the polar angle of the points (1, 1), (−1, 1), (1, −1) and (3, −1)? See figure 0.5.

3Some authors (like Stewart) extend the definition to the case in whichr may be negative.

(1, 1) (−1, 1)

(3, −1)

0 Re-axis

Im-axis

ϕ1

ϕ2

ϕ4

1 1

Figure 0.5: The polar angle of the points (1, 1), (−1, 1) and (3, −1).

The point (1, 1) lies in the first quadrant, so the polar angle is ϕ1 = arctan¹₁ = ^π₄.

The point (−1, 1) lies in the second quadrant, so the polar angle is ϕ₂ = π+arctan₋₁¹ = ^3π₄ . The point (1, −1) lies in the fourth quadrant, so the polar angle is ϕ3 = arctan⁻¹₁ =−^π₄. The point (3, −1) lies in the fourth quadrant, so the polar angle is ϕ₄ = arctan⁻¹₃ =

− arctan¹₃.

POLAR COORDINATES IN THE COMPLEX PLANE We can apply polar coordinates in the complex plane. In fact, from a = r cos ϕ and b = r sin ϕ we see

z = a + bi = r(cos ϕ + i sin ϕ) where

r =

a²+ b²=|z| and (for a = 0) : tan ϕ = b a

In this situation the polar radius is equal to the modulus of z and now the polar angle ϕ is called the argument of z. Writing complex numbers in polar coordinates, we can describe the complex multiplication as follows. Using the addition formulas for the sine and cosine, we find

zw = |z| |w|(cos ϕ + i sin ϕ)(cos ψ + i sin ψ)

=|z| |w|

(cos ϕ cos ψ − sin ϕ sin ψ) + i(sin ϕ cos ψ + cos ϕ sin ψ)

=|z| |w|

cos(ϕ + ψ) + i sin(ϕ + ψ) .

Thus we have proved the first part of the following theorem. The second part can be proved similarly.

Theorem 0.4. Suppose z and w are two complex numbers, with arguments ϕ and ψ respectively:

z = |z| (cos ϕ + i sin ϕ) w = |w| (cos ψ + i sin ψ) . Then

zw = |z| |w|

cos(ϕ + ψ) + i sin(ϕ + ψ) and for w = 0:

wz = _|w|^|z|

cos(ϕ − ψ) + i sin(ϕ − ψ)

From this theorem we see that to get the product zw we have to multiply the moduli and to add the arguments.

w z zw

0 Re-axis

Im-axis

ϕ ψ ϕ + ψ 1 i

Figure 0.6: The complex multiplication Example 0.5. Let z =√

3 + i and w = 1 + i√

3. Then zw = (√

3 + i)(1 + i√

3) = 4i.

In polar coordinates we can write z and w as z = |z| (cos ϕ + i sin ϕ) = 2

cos^π₆ + i sin^π₆ w = |w| (cos ψ + i sin ψ) = 2

cos^π₃ + i sin^π₃ . On the other hand

zw = 4i = 4

cos^π₂ + i sin^π₂ .

This is in accordance with theorem 0.4, since 2· 2 = 4 and ^π₆ +^π₃ = ^π₂.

THE EXPONENTIAL FUNCTION We have seen that each complex number z can be written as

z = |z| (cos ϕ + i sin ϕ) ,

with ϕ an argument of z. To abbreviate the expression cos ϕ + i sin ϕ we introduce the following notation.

Notation 0.6. Let t be a real number. We will denote the complex number cos t + i sin t as e^it:

e^it= cos t + i sin t (Euler’s formula) Although this notation may seem a little odd, the parts (b) and (c) of the following theorem will give a strong motivation for this definition. A close connection exists between the sine and cosine on one hand and the exponential function on the other hand, but at this moment it is too early to handle this⁴. Note that, by i0 = 0, the notation for t = 0 implies the well known equality e⁰= 1.

Theorem 0.7. For all t ∈^R, s ∈^R: (a) |e^it| = 1;

4Cf. module 2, Taylor series.

(b) e^it· e^is = e^i(t+s);

(c) For each natural number n: (e^it)ⁿ= e^int. (De Moivre’s Theorem.) Proof. (a) |e^it| =

cos²t + sin²t = 1.

(b) This follows immediately from theorem 0.4.

(c) It is clear for n = 1. For n = 2 this follows from (b) by taking s equal to t. For n = 3 we can derive this from (b) (take s = 2t) and the case for n = 2 which we already found. Repeating this we get (c) for all n = 1, 2, . . . .

Remark 0.8. From (b) you can easily find again⁵ the addition formulas for the cosine and sine. On one hand we have

e^i(ϕ+ψ)= cos(ϕ + ψ) + i sin(ϕ + ψ) while on the other hand

e^i(ϕ+ψ)= e^iϕe^iψ= (cos ϕ + i sin ϕ)(cos ψ + i sin ψ))

= (cos ϕ cos ψ − sin ϕ sin ψ) + i(cos ϕ sin ψ + cos ψ sin ϕ).

Equating the real and imaginary parts implies the addition formulas.

The new notation gives the following short formula for a number z with argument ϕ:

z = |z|e^iϕ

Example 0.9. The complex numbers with modulus 1 form together the unit circle in the complex plane, that is the circle with center O and radius 1.

Each number z of the form e^it lies on this circle: this follows from theorem 0.7 (a). On the other hand each number z on the unit circle can be written in the form e^it. For if ϕ is an argument for z, then z = |z|e^iϕ= e^iϕ. So we can choose t = ϕ. See figure 0.7.

We illustrate this by the the points 1, i, −1 en −i on the unit circle. We have 1 = · · · = e^−4πi = e^−2πi = e^0πi = e^2πi= e^4πi = . . .

and likewise

i = · · · = e^−32πi= e^12πi= e^52πi= . . .

−1 = · · · = e^−πi = e^πi= e^3πi = . . .

−i = · · · = e^−12πi= e^32πi= e^72πi= . . . , and so on.

Now we can extend the definition of the exponential function to all complex numbers: let z = a + bi be a complex number, then we define e^z by:

e^z = e^ae^ib For instance e^−1+iπ/2= i/e, since

e^−1+iπ/2= e⁻¹e^iπ/2= ¹_e(cos^π₂ + i sin^π₂) = ¹_e(0 + i) = i/e.

5This is not a proof of the addition formulas, only a way to recollect them

i^π= -1

π π= i

i

iϕ

e^{i 0}

e = 1

0

ϕ

e^-i = -i

Re-axis Im-axis

e e

2 2

Figure 0.7: The unit circle in the complex plane

EQUATIONS The quadratic equation ax²+ bx + c = 0 with real coefficients has as its solutions

x = −b ±√

b²− 4ac

2a .

Using complex numbers, this formula is also valid for b²− 4ac < 0.

Example 0.10. The solutions of the equation x²+ x + 1 = 0 are x = −1 ±√

1²− 4 · 1

2 = −1 ±√

−3

2 = −1 ±√

3 i

2 .

In particular we see that each quadratic equation with real coefficients has at least one solution. In general we know that each equation of the form

a_nxⁿ+ a_n−1xⁿ⁻¹+· · · + a₁x + a₀ = 0

in which n ≥ 1 and a₀, . . . , a_n are complex numbers and in which z is the unknown, has always at least one solution (and at most n) in the complex numbers. This fact is known as the Principal theorem of the Algebra . We will not prove this theorem here.

For n = 1 and n = 2 there are relatively simple formulas for the solution of this equation, but for n ≥ 3 the solutions are usually approximated with computer programs, such as Maple.

Another equation that can easily be solved is the n-th degree equation (n ≥ 1) zⁿ= w

in which w is an arbitrary complex number and z is to be solved. For w = 0 we see immediately that z = 0 is the only solution. Therefore we assume from now that w = 0.

If we write w and z in polar coordinates, w = |w|e^iϕ and z = re^iψ, we can rewrite the equation as

rⁿe^inψ =|w|e^iϕ.

The unknown variables r and ψ have to be solved from this equation. This is equivalent with solving the system of equations

rⁿ=|w|

nψ = ϕ + 2kπ with k ∈^Z.

Example 0.11. 1. The equation z² = i has exactly two complex solutions: z =

±¹₂√

2(1 + i). To derive this we write i = e^iπ2 and we find r² = 1 and 2ψ = ^π₂ + 2kπ, so that r = 1 and ψ = ^π₄ or ψ = ^−3π₄ (or an angle differing an integer multiple of 2π from one of these). This gives the solutions:

z = e^iπ4 = ¹₂√

2(1 + i) and z = e^i−3π4 =−¹₂√

2(1 + i).

2. In the same way we can solve the equation z⁵ = −1 − i. Since −1 − i =√

2e^5πi/4, we find as solutions for z:

10√

2 e^iπ4, ¹⁰√

2 e^i(π4+2π15), ¹⁰√

2 e^i(π4+2π25), ¹⁰√

2 e^i(π4+2π35), ¹⁰√

2 e^i(π4+2π45). 3. The equation zⁿ= 1 has exactly n complex solutions:

z = e^i2kπn , k = 0, . . . , n − 1.

These solutions are called the n-th power unit roots or n-th roots of unity . In the complex plane they form a regular n-polygon, with all corners on the unit circle; see figure 0.8.

For n = 2 we get the solutions z = eⁱ⁰ = 1 and z = e^iπ = −1, which was to be expected.

Im-axis

1

π

0

/3

Figure 0.8: The 6th roots of unity

In the preceding we have computed all complex solutions of the equations zⁿ− w = 0 and az²+ bz + c = 0 (a, b, c ∈^R).

The expressions zⁿ− w and az²+ bz + c are special cases of polynomials in the variable z:

Definition 0.12. A polynomial in the complex variable z is an expression p(z) of the form

p(z) = anzⁿ+ an−1zⁿ⁻¹+· · · + a1z + a0.

The numbers a0, a1, . . . , an are complex and we assume that an = 0. These numbers are called the coefficients and n the degree af the polynomial. Solutions of the equation p(z) = 0 are called roots of the equation or zeros of the polynomial p(z). If we want to emphasize that the variable z and the coefficients ak are complex, we call the polynomial a complex polynomial.

Example 0.13. The polynomial z²− 2z + (1 − i) is a polynomial in z of degree 2 with coefficients a₂ = 1, a₁=−2 and a₀ = 1− i.

The polynomial zⁿ− c is a polynomial in z of degree n with coefficients an = 1, an−1 =

· · · = a₁ = 0 and a₀ =−c.

What is the way to determine the zeros of an arbitrary polynomial? In any case a solution has to exist (Principal theorem of the Algebra), but often there is no nice expression. If we are lucky, it is easy, namely when we can recognize one or more of the solutions.

Example 0.14 (Long division). We will try to solve the 3rd degree equation

z³+ z²+ 2z + 2 = 0.

By trial and error, you can see that z = −1 is a solution. Now z = −1 is a zero of the polynomial z + 1 and we can carry out the division below:

z³+ z²+ 2z + 2 z²× (z + 1) = z³+ z² −

2z + 2 2× (z + 1) = 2z + 2 −

0 The result is that

z³+ z²+ 2z + 2 = (z²+ 2)(z + 1).

We conclude that

z³+ z²+ 2z + 2 = 0 ⇐⇒ (z²+ 2)(z + 1) = 0

⇐⇒ z² =−2 of z = −1

⇐⇒ z = i√

2 of z = −i√

2 of z = −1.

In this context we say that we have divided out the factor z + 1. To get all solutions we only had to find the zeros of z²+ 2.

If we have found a zero (for instance by trying some numbers), then this division always works. If p(z) is a polynomial of degree n ≥ 2 and we divide by z − α, then we get a polynomial q(z), with a constant c as a remainder, so p(z) = (z − α)q(z) + c. However, if α is a zero of p(z), we see, by substituting α for z, that c has to zero. Therefore we have:

Theorem 0.15. Let p(z) be a polynomial of degree n ≥ 2. If α ∈ ^C is a zero of p(z), then there exists a polynomial q(z) of degree n − 1 such that

p(z) = (z − α)q(z) for all z ∈^C.

PARTIAL FRACTIONS In some ways integer numbers and polynomials look alike.

We have seen that polynomials can split into factors just like integers and that you can divide one polynomial by another. Now you can make a fraction of polynomials, with both numerator and denominator a polynomial (a so called rational function). Two of these fractions can be added by making a common denominator, as in:

x

x²+ 1− 1

x − 1= x(x − 1)

(x²+ 1)(x − 1) − x²+ 1

(x²+ 1)(x − 1) =

= x(x − 1) − (x²+ 1)

(x²+ 1)(x − 1) =− x + 1 (x²+ 1)(x − 1).

Frequently we need the reverse procedure en we have to split such a fraction with composite denominator into separate fractions with simpler denominators. There is a general method to accomplish this (the so called partial fraction decomposition, see also Stewart 7.4).

We don’t treat this here in full, but we only give a simple example. Let h be the rational function defined by

h(x) = 2x − 1 (2− x)(1 + x). We will try to find real constants A and B such that

2x − 1

(2− x)(1 + x) = A

(2− x) + B (1 + x). Making the right hand side to one fraction, we find

A

(2− x)+ B

(1 + x) = A(1 + x) + B(2 − x)

(2− x)(1 + x) = (A + 2B) + x(A − B) (2− x)(1 + x) . This function is exactly equal to _(2−x)(1+x)^2x−1 if

 A + 2B = −1 A − B = 2.

This system has as its solution

A = −B = 1, so we conclude that

h(x) = 2x − 1

(2− x)(1 + x) = 1

(2− x)− 1 (1 + x).

Notice that in this case the denominator of h(x) has two factors of degree 1 and that the degree of the numerator (which is 1) is smaller than the degree of the denominator. In such a case it is always possible to perform such a decomposition into partial fractions like the above.

(1) (2 − 2i) − (2 + i), (2) (−1 + 2i)(1 − 3i), (3) 2 + 3i

1 − 5i,

(4) 4

4 + 3i, (5) e²⁺³ⁱ.

Ex. II. Find the complex conjugate, the modulus and the argument of the follow- ing numbers and write them in the form |z| e^iϕ.

(6) √

3 − i, (7) −4i, (8) −3 − 3i, (9) −1 + i, (10) (1 + i)²⁰.

Ex. III. Solve the following equations.

(11) z²+ 8z + 17 = 0, (12) z⁵= 32,

(13) z⁶= −i.

Ex. IV. Determine a partial fraction decomposition of:

(14) x + 5 x²+ x − 2,

(15) 1

x³+ x, (16) x⁴+ 1

x⁴− 1.