5.5 Modulus of a complex number

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DEFINITION 5.5.1 (Modulus) If z = x + iy, the modulus of z is the

non{negative real number jzj de¯ned by jzj =

p

x2 + y2. Geometrically, the

modulus of z is the distance from z to 0 (see Figure 5.3).

More generally, jz1¡z2j is the distance between z1 and z2 in the complex

plane. For

jz1 ¡ z2j = j(x1 + iy1) ¡ (x2 + iy2)j = j(x1 ¡ x2) + i(y1 ¡ y2)j

=

p

(x1 ¡ x2)2 + (y1 ¡ y2)2:

The following properties of the modulus are easy to verify, using the identity

jzj2 = zz:

(i) jz1z2j = jz1jjz2j;

100 CHAPTER 5. COMPLEX NUMBERS

(ii) jz¡1j = jzj¡1;

(iii)

¯¯¯¯

z1

z2

¯¯¯¯

= jz1j

jz2j

.

For example, to prove (i):

jz1z2j2 = (z1z2)z1z2 = (z1z2)z1 z2

= (z1z1)(z2z2) = jz1j2jz2j2 = (jz1jjz2j)2:

Hence jz1z2j = jz1jjz2j.

EXAMPLE 5.5.1 Find jzj when z =

(1 + i)4

(1 + 6i)(2 ¡ 7i)

.

Solution.

jzj = j1 + ij4

j1 + 6ijj2 ¡ 7ij

=

(p12 + 12)4

p12 + 62

p

22 + (¡7)2

=

4

p37p53

:

THEOREM 5.5.1 (Ratio formulae) If z lies on the line through z1 and

z2:

z = (1 ¡ t)z1 + tz2; t 2 R;

we have the useful ratio formulae:

(i)

¯¯¯¯

z ¡ z1

z ¡ z2

¯¯¯¯

=

¯¯¯¯

t

1 ¡ t

¯¯¯¯

if z 6= z2,

(ii)

¯¯¯¯

z ¡ z1

z1 ¡ z2

¯¯¯¯

= jtj.

Circle equations. The equation jz ¡ z0j = r, where z0 2 C and r >

0, represents the circle centre z0 and radius r. For example the equation

jz ¡ (1 + 2i)j = 3 represents the circle (x ¡ 1)2 + (y ¡ 2)2 = 9.

Another useful circle equation is the circle of Apollonius :

¯¯¯¯

z ¡ a

z ¡ b

¯¯¯¯

= ¸;

5.5. MODULUS OF A COMPLEX NUMBER 101

¾ -

6

?

x

y

Figure 5.4: Apollonius circles: jz+2ij

jz¡2ij

= 1

4 ; 3

8 ; 1

2 ; 5

8 ; 4

1 ; 8

3 ; 2

1 ; 8

5 .

where a and b are distinct complex numbers and ¸ is a positive real number,

¸ 6= 1. (If ¸ = 1, the above equation represents the perpendicular bisector

of the segment joining a and b.)

An algebraic proof that the above equation represents a circle, runs as

follows. We use the following identities:

(i) jz ¡ aj2 = jzj2 ¡ 2Re (za) + jaj2

(ii) Re (z1 § z2) = Re z1 § Re z2

(iii) Re (tz) = tRe z if t 2 R.

We have

¯¯¯¯

z ¡ a

z ¡ b

¯¯¯¯

= ¸ , jz ¡ aj2 = ¸2jz ¡ bj2

, jzj2 ¡ 2Re fzag + jaj2 = ¸2(jzj2 ¡ 2Re fzbg + jbj2)

, (1 ¡ ¸2)jzj2 ¡ 2Re fz(a ¡ ¸2b)g = ¸2jbj2 ¡ jaj2

, jzj2 ¡ 2Re

½

z

µ

a ¡ ¸2b

1 ¡ ¸2

¶¾

=

¸2jbj2 ¡ jaj2

1 ¡ ¸2

, jzj2 ¡ 2Re

½

z

µ

a ¡ ¸2b

1 ¡ ¸2

¶¾

+

¯¯¯¯

a ¡ ¸2b

1 ¡ ¸2

¯¯¯¯

2

=

¸2jbj2 ¡ jaj2

1 ¡ ¸2 +

¯¯¯¯

a ¡ ¸2b

1 ¡ ¸2

¯¯¯¯

2

:

102 CHAPTER 5. COMPLEX NUMBERS

Now it is easily veri¯ed that

ja ¡ ¸2bj2 + (1 ¡ ¸2)(¸2jbj2 ¡ jaj2) = ¸2ja ¡ bj2:

So we obtain

¯¯¯¯

z ¡ a

z ¡ b

¯¯¯¯

= ¸ ,

¯¯¯¯

z ¡

µ

a ¡ ¸2b

1 ¡ ¸2

¶¯¯¯¯

2

=

¸2ja ¡ bj2

j1 ¡ ¸2j2

,

¯¯¯¯

z ¡

µ

a ¡ ¸2b

1 ¡ ¸2

¶¯¯¯¯ =

¸ja ¡ bj

j1 ¡ ¸2j

:

The last equation represents a circle centre z0, radius r, where

z0 =

a ¡ ¸2b

1 ¡ ¸2 and r =

¸ja ¡ bj

j1 ¡ ¸2j

:

There are two special points on the circle of Apollonius, the points z1 and

z2 de¯ned by

z1 ¡ a

z1 ¡ b

= ¸ and

z2 ¡ a

z2 ¡ b

= ¡¸;

or

z1 =

a ¡ ¸b

1 ¡ ¸

and z2 =

a + ¸b

1 + ¸

: (5.3)

It is easy to verify that z1 and z2 are distinct points on the line through a

and b and that z0 = z1+z2

2 . Hence the circle of Apollonius is the circle based

on the segment z1; z2 as diameter.

EXAMPLE 5.5.2 Find the centre and radius of the circle

jz ¡ 1 ¡ ij = 2jz ¡ 5 ¡ 2ij.

Solution. Method 1. Proceed algebraically and simplify the equation

jx + iy ¡ 1 ¡ ij = 2jx + iy ¡ 5 ¡ 2ij

or

jx ¡ 1 + i(y ¡ 1)j = 2jx ¡ 5 + i(y ¡ 2)j:

Squaring both sides gives

(x ¡ 1)2 + (y ¡ 1)2 = 4((x ¡ 5)2 + (y ¡ 2)2);

which reduces to the circle equation

x2 + y2 ¡

38

3

x ¡

14

3

y + 38 = 0:

5.6. ARGUMENT OF A COMPLEX NUMBER 103

Completing the square gives

(x ¡

19

3

)2 + (y ¡

7

3

)2 =

µ

19

3

¶2

+

µ

7

3

¶2

¡ 38 =

68

9

;

so the centre is ( 19

3 ; 7

3 ) and the radius is

q

68

9 .

Method 2. Calculate the diametrical points z1 and z2 de¯ned above by

equations 5.3:

z1 ¡ 1 ¡ i = 2(z1 ¡ 5 ¡ 2i)

z2 ¡ 1 ¡ i = ¡2(z2 ¡ 5 ¡ 2i):

We ¯nd z1 = 9 + 3i and z2 = (11 + 5i)=3. Hence the centre z0 is given by

z0 =

z1 + z2

2

=

19

3

+

7

3

i

and the radius r is given by

r = jz1 ¡ z0j =

¯¯¯¯

µ

19

3

+

7

3

i

¡ (9 + 3i)

¯¯¯¯

=

¯¯¯¯

¡

8

3 ¡

2

3

i

¯¯¯¯

=

p68

3

:

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