5.2 Calculating with complex numbers

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We can now do all the standard linear algebra calculations over the ¯eld of

complex numbers { ¯nd the reduced row{echelon form of an matrix whose el-

ements are complex numbers, solve systems of linear equations, ¯nd inverses

and calculate determinants.

For example,

¯¯¯¯

1 + i 2 ¡ i

7 8 ¡ 2i

¯¯¯¯

= (1 + i)(8 ¡ 2i) ¡ 7(2 ¡ i)

= (8 ¡ 2i) + i(8 ¡ 2i) ¡ 14 + 7i

= ¡4 + 13i 6= 0:

92 CHAPTER 5. COMPLEX NUMBERS

Then by Cramer's rule, the linear system

(1 + i)z + (2 ¡ i)w = 2 + 7i

7z + (8 ¡ 2i)w = 4 ¡ 9i

has the unique solution

z =

¯¯¯¯

2 + 7i 2 ¡ i

4 ¡ 9i 8 ¡ 2i

¯¯¯¯

¡4 + 13i

=

(2 + 7i)(8 ¡ 2i) ¡ (4 ¡ 9i)(2 ¡ i)

¡4 + 13i

=

2(8 ¡ 2i) + (7i)(8 ¡ 2i) ¡ f(4(2 ¡ i) ¡ 9i(2 ¡ i)g

¡4 + 13i

=

16 ¡ 4i + 56i ¡ 14i2 ¡ f8 ¡ 4i ¡ 18i + 9i2g

¡4 + 13i

=

31 + 74i

¡4 + 13i

=

(31 + 74i)(¡4 ¡ 13i)

(¡4)2 + 132

=

838 ¡ 699i

(¡4)2 + 132

=

838

185 ¡

699

185

i

and similarly w = ¡698

185

+

229

185

i.

An important property enjoyed by complex numbers is that every com-

plex number has a square root:

THEOREM 5.2.1

If w is a non{zero complex number, then the equation z2 = w has precisely

two solutions z 2 C.

Proof. Case 1. Suppose b = 0. Then if a > 0, z = pa is a solution, while

if a < 0, ip¡a is a solution.

Case 2. Suppose b 6= 0. Let z = x+iy; w = a+ib; x; y; a; b 2 R. Then

the equation z2 = w becomes

(x + iy)2 = x2 ¡ y2 + 2xyi = a + ib;

5.2. CALCULATING WITH COMPLEX NUMBERS 93

so equating real and imaginary parts gives

x2 ¡ y2 = a and 2xy = b:

Hence x 6= 0 and y = b=(2x). Consequently

x2 ¡

µ

b

2x

2 = a;

so 4x4 ¡ 4ax2 ¡ b2 = 0 and 4(x2)2 ¡ 4a(x2) ¡ b2 = 0. Hence

x2 =

4a § p16a2 + 16b2

8

=

a § pa2 + b2

2

:

However x2 > 0, so we must take the + sign, as a ¡ pa2 + b2 < 0. Hence

x2 =

a + pa2 + b2

2

; x = §

s

a + pa2 + b2

2

:

Then y is determined by y = b=(2x).

EXAMPLE 5.2.1 Solve the equation z2 = 1 + i.

Solution. Put z = x + iy. Then the equation becomes

(x + iy)2 = x2 ¡ y2 + 2xyi = 1 + i;

so equating real and imaginary parts gives

x2 ¡ y2 = 1 and 2xy = 1:

Hence x 6= 0 and y = b=(2x). Consequently

x2 ¡

µ

1

2x

2 = 1;

so 4x4 ¡ 4x2 ¡ 1 = 0. Hence

x2 =

4 § p16 + 16

8

=

1 § p2

2

:

Hence

x2 =

1 + p2

2

and x = §

s

1 + p2

2

:

94 CHAPTER 5. COMPLEX NUMBERS

Then

y =

1

2x

= §

1

p2

p

1 + p2

:

Hence the solutions are

z = §

0

@

s

1 + p2

2

+

i

p2

p

1 + p2

1

A:

EXAMPLE 5.2.2 Solve the equation z2 + (p3 + i)z + 1 = 0.

Solution. Because every complex number has a square root, the familiar

formula

z = ¡b § pb2 ¡ 4ac

2a

for the solution of the general quadratic equation az2 + bz + c = 0 can be

used, where now a(6= 0); b; c 2 C. Hence

z = ¡(p3 + i) §

q

(p3 + i)2 ¡ 4

2

= ¡(p3 + i) §

q

(3 + 2p3i ¡ 1) ¡ 4

2

= ¡(p3 + i) §

p

¡2 + 2p3i

2

:

Now we have to solve w2 = ¡2 + 2p3i. Put w = x + iy. Then w2 =

x2 ¡ y2 + 2xyi = ¡2 + 2p3i and equating real and imaginary parts gives

x2 ¡ y2 = ¡2 and 2xy = 2p3. Hence y = p3=x and so x2 ¡ 3=x2 = ¡2. So

x4 + 2x2 ¡ 3 = 0 and (x2 + 3)(x2 ¡ 1) = 0. Hence x2 ¡ 1 = 0 and x = §1.

Then y = §p3. Hence (1 + p3i)2 = ¡2 + 2p3i and the formula for z now

becomes

z = ¡p3 ¡ i § (1 + p3i)

2

=

1 ¡ p3 + (1 + p3)i

2

or ¡1 ¡ p3 ¡ (1 + p3)i

2

:

EXAMPLE 5.2.3 Find the cube roots of 1.

5.3. GEOMETRIC REPRESENTATION OF C 95

Solution. We have to solve the equation z3 = 1, or z3 ¡ 1 = 0. Now

z3 ¡ 1 = (z ¡ 1)(z2 + z + 1). So z3 ¡ 1 = 0 ) z ¡ 1 = 0 or z2 + z + 1 = 0.

But

z2 + z + 1 = 0 ) z = ¡1 § p12 ¡ 4

2

= ¡1 § p3i

2

:

So there are 3 cube roots of 1, namely 1 and (¡1 § p3i)=2.

We state the next theorem without proof. It states that every non{

constant polynomial with complex number coe±cients has a root in the

¯eld of complex numbers.

THEOREM 5.2.2 (Gauss) If f(z) = anzn + an¡1zn¡1 + ¢ ¢ ¢ + a1z + a0,

where an 6= 0 and n ¸ 1, then f(z) = 0 for some z 2 C.

It follows that in view of the factor theorem, which states that if a 2 F is

a root of a polynomial f(z) with coe±cients from a ¯eld F, then z ¡ a is a

factor of f(z), that is f(z) = (z ¡ a)g(z), where the coe±cients of g(z) also

belong to F. By repeated application of this result, we can factorize any

polynomial with complex coe±cients into a product of linear factors with

complex coe±cients:

f(z) = an(z ¡ z1)(z ¡ z2) ¢ ¢ ¢ (z ¡ zn):

There are available a number of computational algorithms for ¯nding good

approximations to the roots of a polynomial with complex coe±cients.

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