argument of product and quotient

argument of product and quotient

Using the

distributive law

, we perform the

multiplication

(

cos

⁡

φ

1

+

i

⁢

sin

⁡

φ

1

)

⁢

(

cos

⁡

φ

2

+

i

⁢

sin

⁡

φ

2

)

=

(

cos

⁡

φ

1

⁢

cos

⁡

φ

2

-

sin

⁡

φ

1

⁢

sin

⁡

φ

2

)

+

i

⁢

(

sin

⁡

φ

1

⁢

cos

⁡

φ

2

+

cos

⁡

φ

1

⁢

sin

⁡

φ

2

)

.

Using the

addition formulas

of cosine (

http://planetmath.org/GoniometricFormulae

) and sine (

http://planetmath.org/GoniometricFormulae

) we still obtain the formula

(

cos

⁡

φ

1

+

i

⁢

sin

⁡

φ

1

)

⁢

(

cos

⁡

φ

2

+

i

⁢

sin

⁡

φ

2

)

=

cos

⁡

(

φ

1

+

φ

2

)

+

i

⁢

sin

⁡

(

φ

1

+

φ

2

)

.

(1)

The

inverse number

of

cos

⁡

φ

2

+

i

⁢

sin

⁡

φ

2

is calculated as follows:

1

cos

⁡

φ

2

+

i

⁢

sin

⁡

φ

2

=

cos

⁡

φ

2

-

i

⁢

sin

⁡

φ

2

(

cos

⁡

φ

2

-

i

⁢

sin

⁡

φ

2

)

⁢

(

cos

⁡

φ

2

+

i

⁢

sin

⁡

φ

2

)

=

cos

⁡

φ

2

-

i

⁢

sin

⁡

φ

2

cos

2

⁡

φ

2

+

sin

2

⁡

φ

2

This equals

cos

⁡

φ

2

-

i

⁢

sin

⁡

φ

2

,  and since the cosine is an even (

http://planetmath.org/EvenFunction

) and the sine an

odd function

, we have

1

cos

⁡

φ

2

+

i

⁢

sin

⁡

φ

2

=

cos

⁡

(

-

φ

2

)

+

i

⁢

sin

⁡

(

-

φ

2

)

.

(2)

The equations (1) and (2) imply

cos

⁡

φ

1

+

i

⁢

sin

⁡

φ

1

cos

⁡

φ

2

+

i

⁢

sin

⁡

φ

2

=

(

cos

⁡

φ

1

+

i

⁢

sin

⁡

φ

1

)

⁢

(

cos

⁡

(

-

φ

2

)

+

i

⁢

sin

⁡

(

-

φ

2

)

)

=

cos

⁡

(

φ

1

+

(

-

φ

2

)

)

+

i

⁢

sin

⁡

(

φ

1

+

(

-

φ

2

)

)

,

i.e.

cos

⁡

φ

1

+

i

⁢

sin

⁡

φ

1

cos

⁡

φ

2

+

i

⁢

sin

⁡

φ

2

=

cos

⁡

(

φ

1

-

φ

2

)

+

i

⁢

sin

⁡

(

φ

1

-

φ

2

)

.

(3)

According to the formulae (1) and (3), for the

complex numbers

z

1

=

r

1

⁢

(

cos

⁡

φ

1

+

i

⁢

sin

⁡

φ

1

)

⁢

and

⁢

z

2

=

r

2

⁢

(

cos

⁡

φ

2

+

i

⁢

sin

⁡

φ

2

)

we have

z

1

⁢

z

2

=

r

1

⁢

r

2

⁢

(

cos

⁡

(

φ

1

+

φ

2

)

+

i

⁢

sin

⁡

(

φ

1

+

φ

2

)

)

,

z

1

z

2

=

r

1

r

2

⁢

(

cos

⁡

(

φ

1

-

φ

2

)

+

i

⁢

sin

⁡

(

φ

1

-

φ

2

)

)

.

Thus we have the

Theorem

.

The

modulus

of the product of two complex numbers equals the product of the moduli of the factors and the

argument

equals the sum of the arguments of the factors (

http://planetmath.org/Product

).  The modulus of the

quotient

of two complex numbers equals the quotient of the moduli of the

dividend

and the

divisor

and the argument equals the

difference

of the arguments of the dividend and the divisor.

Remark.

The equation (1) may be by induction generalised for more than two factors of the left hand ; then the special case where all factors are equal gives

de Moivre identity

.

Example.

Since

(

2

+

i

)

⁢

(

3

+

i

)

=

5

+

5

⁢

i

=

5

⁢

e

π

4

,

one has

arctan

⁡

1

2

+

arctan

⁡

1

3

=

π

4

.

Title

argument of product and quotient

Canonical

name

ArgumentOfProductAndQuotient

Date of creation

2013-03-22 17:45:20

Last modified on

2013-03-22 17:45:20

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Theorem

Classification

msc 30-00

Classification

msc 26A09

Synonym

product and quotient of complex numbers

Related topic

Argument

Related topic

PolarCoordinates

Related topic

ModulusOfComplexNumber

Related topic

Complex

Related topic

EqualityOfComplexNumbers

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