5.6 Argument of a complex number

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Let z = x + iy be a non{zero complex number, r = jzj =

p

x2 + y2. Then

we have x = r cos µ; y = r sin µ, where µ is the angle made by z with the

positive x{axis. So µ is unique up to addition of a multiple of 2¼ radians.

DEFINITION 5.6.1 (Argument) Any number µ satisfying the above

pair of equations is called an argument of z and is denoted by arg z. The

particular argument of z lying in the range ¡¼ < µ · ¼ is called the principal

argument of z and is denoted by Arg z (see Figure 5.5).

We have z = r cos µ + ir sin µ = r(cos µ + i sin µ) and this representation

of z is called the polar representation or modulus{argument form of z.

EXAMPLE 5.6.1 Arg 1 = 0, Arg (¡1) = ¼, Arg i = ¼

2 , Arg (¡i) = ¡¼

2 .

We note that y=x = tan µ if x 6= 0, so µ is determined by this equation up

to a multiple of ¼. In fact

Arg z = tan¡1 y

x

+ k¼;

104 CHAPTER 5. COMPLEX NUMBERS

¾ -

6

?

½

½

½

½>

x

r y

z

µ

Figure 5.5: The argument of z: arg z = µ.

where k = 0 if x > 0; k = 1 if x < 0; y > 0; k = ¡1 if x < 0; y < 0.

To determine Arg z graphically, it is simplest to draw the triangle formed

by the points 0; x; z on the complex plane, mark in the positive acute angle

® between the rays 0; x and 0; z and determine Arg z geometrically, using

the fact that ® = tan¡1(jyj=jxj), as in the following examples:

EXAMPLE 5.6.2 Determine the principal argument of z for the followig

complex numbers:

z = 4 + 3i; ¡4 + 3i; ¡4 ¡ 3i; 4 ¡ 3i:

Solution. Referring to Figure 5.6, we see that Arg z has the values

®; ¼ ¡ ®; ¡¼ + ®; ¡®;

where ® = tan¡1 3

4 .

An important property of the argument of a complex number states that

the sum of the arguments of two non{zero complex numbers is an argument

of their product:

THEOREM 5.6.1 If µ1 and µ2 are arguments of z1 and z2, then µ1 + µ2

is an argument of z1z2.

Proof. Let z1 and z2 have polar representations z1 = r1(cos µ1 + i sin µ1)

and z2 = r2(cos µ2 + i sin µ2). Then

z1z2 = r1(cos µ1 + i sin µ1)r2(cos µ2 + i sin µ2)

= r1r2(cos µ1 cos µ2 ¡ sin µ1 sin µ2 + i(cos µ1 sin µ2 + sin µ1 cos µ2))

= r1r2(cos (µ1 + µ2) + i sin (µ1 + µ2));

5.6. ARGUMENT OF A COMPLEX NUMBER 105

¾ -

6

?

x

y

4 + 3i

½®

½

½

½>

¾ -

6

?

x

y

¡4 + 3i

®Z

Z

Z

Z}

¾ -

6

?

x

y

¡4 ¡ 3i

®½

½

½

½=

¾ -

6

?

x

y

4 ¡ 3i

Z®

Z

Z

Z~

Figure 5.6: Argument examples.

which is the polar representation of z1z2, as r1r2 = jz1jjz2j = jz1z2j. Hence

µ1 + µ2 is an argument of z1z2.

An easy induction gives the following generalization to a product of n

complex numbers:

COROLLARY 5.6.1 If µ1; : : : ; µn are arguments for z1; : : : ; zn respectively,

then µ1 + ¢ ¢ ¢ + µn is an argument for z1 ¢ ¢ ¢ zn.

Taking µ1 = ¢ ¢ ¢ = µn = µ in the previous corollary gives

COROLLARY 5.6.2 If µ is an argument of z, then nµ is an argument for

zn.

THEOREM 5.6.2 If µ is an argument of the non{zero complex number

z, then ¡µ is an argument of z¡1.

Proof. Let µ be an argument of z. Then z = r(cos µ+i sin µ), where r = jzj.

Hence

z¡1 = r¡1(cos µ + i sin µ)¡1

= r¡1(cos µ ¡ i sin µ)

= r¡1(cos(¡µ) + i sin(¡µ)):

106 CHAPTER 5. COMPLEX NUMBERS

Now r¡1 = jzj¡1 = jz¡1j, so ¡µ is an argument of z¡1.

COROLLARY 5.6.3 If µ1 and µ2 are arguments of z1 and z2, then µ1¡µ2

is an argument of z1=z2.

In terms of principal arguments, we have the following equations:

(i) Arg (z1z2) = Arg z1+Arg z2 + 2k1¼,

(ii) Arg (z¡1) = ¡Arg z + 2k2¼,

(iii) Arg (z1=z2) = Arg z1¡Arg z2 + 2k3¼,

(iv) Arg (z1 ¢ ¢ ¢ zn) = Arg z1 + ¢ ¢ ¢ +Arg zn + 2k4¼,

(v) Arg (zn) = n Arg z + 2k5¼,

where k1; k2; k3; k4; k5 are integers.

In numerical examples, we can write (i), for example, as

Arg (z1z2) ´ Arg z1 + Arg z2:

EXAMPLE 5.6.3 Find the modulus and principal argument of

z =

Ãp3 + i

1 + i

!17

and hence express z in modulus{argument form.

Solution. jzj = jp3 + ij17

j1 + ij17 =

217

(p2)17

= 217=2.

Arg z ´ 17Arg

Ãp3 + i

1 + i

!

= 17(Arg (p3 + i) ¡ Arg (1 + i))

= 17

³¼

6 ¡

¼

4

´

= ¡17¼

12

:

Hence Arg z =

¡

¡17¼

12

¢

+ 2k¼, where k is an integer. We see that k = 1 and

hence Arg z = 7¼

12 . Consequently z = 217=2

¡

cos 7¼

12 + i sin 7¼

12

¢

.

DEFINITION 5.6.2 If µ is a real number, then we de¯ne eiµ by

eiµ = cos µ + i sin µ:

More generally, if z = x + iy, then we de¯ne ez by

ez = exeiy:

5.7. DE MOIVRE'S THEOREM 107

For example,

e

i¼

2 = i; ei¼ = ¡1; e¡i¼

2 = ¡i:

The following properties of the complex exponential function are left as

exercises:

THEOREM 5.6.3 (i) ez1ez2 = ez1+z2 ,

(ii) ez1 ¢ ¢ ¢ ezn = ez1+¢¢¢+zn,

(iii) ez 6= 0,

(iv) (ez)¡1 = e¡z,

(v) ez1=ez2 = ez1¡z2 ,

(vi) ez = ez.

THEOREM 5.6.4 The equation

ez = 1

has the complete solution z = 2k¼i; k 2 Z.

Proof. First we observe that

e2k¼i = cos (2k¼) + i sin (2k¼) = 1:

Conversely, suppose ez = 1; z = x + iy. Then ex(cos y + i sin y) = 1. Hence

ex cos y = 1 and ex sin y = 0. Hence sin y = 0 and so y = n¼; n 2 Z. Then

ex cos (n¼) = 1, so ex(¡1)n = 1, from which follows (¡1)n = 1 as ex > 0.

Hence n = 2k; k 2 Z and ex = 1. Hence x = 0 and z = 2k¼i.

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