complex arithmetic-geometric mean

complex arithmetic-geometric mean

It is also possible to define the

arithmetic-geometric mean

for

complex numbers

. To do this, we first must make the

geometric
mean

unambiguous by choosing a branch of the

square root

. We
may do this as follows: Let

a

and

b

br two non-zero complex
numbers such that

a

≠

s

⁢

b

for any

real number

s

<

0

. Then
we will say that

c

is the geometric mean of

a

and

b

if

c

2

=

a

⁢

b

and

c

is a convex combination of

a

and

b

(i.e.

c

=

s

⁢

a

+

t

⁢

b

for positive real numbers

s

and

t

).

Geometrically, this may be understood as follows: The condition

a

≠

s

⁢

b

means that the angle between

0

⁢

a

and

0

⁢

b

differs
from

π

. The square root of

a

⁢

b

will lie on a line bisecting
this angle, at a distance

|

a

⁢

b

|

from

0

. Our condition
states that we should choose

c

such that

0

⁢

c

bisects the angle
smaller than

π

, as in the figure below:

{xy}

,

(

2

,

-

1

)

*

0

,

(

0

,

0

)

;

(

50

,

50

)

*

*

@

-

;

(

52

,

52

)

*

b

,

(

0

,

0

)

;

(

-

16

,

16

)

*

*

@

-

,

(

-

18

,

18

)

*

a

,

(

0

,

0

)

;

(

0

,

40

)

*

*

@

-

,

(

0

,

42

)

*

c

,

(

0

,

0

)

;

(

0

,

-

40

)

*

*

@

-

,

(

0

,

-

42

)

*

-

c

Analytically, if we pick a polar representation

a

=

|

a

|

⁢

e

i

⁢

α

,

b

=

|

b

|

⁢

e

i

⁢

β

with

|

α

-

β

|

<

π

, then

c

=

|

a

⁢

b

|

⁢

e

i

⁢

α

+

β

2

. Having clarified this preliminary item,
we now proceed to the main

definition

.

As in the real case, we will define

sequences

of geometric and

arithmetic
means

recursively and show that they converge to the same limit. With our
convention, these are defined as follows:

g

0

=

a

a

0

=

b

g

n

+

1

=

a

n

⁢

g

n

a

n

+

1

=

a

n

+

g

n

2

We shall first show that the phases of these sequences converge. As above,
let us define

α

and

β

by the conditions

a

=

|

a

|

⁢

e

i

⁢

α

,

b

=

|

b

|

⁢

e

i

⁢

β

, and

|

α

-

β

|

<

π

. Suppose that

z

and

w

are any two complex numbers such that

z

=

|

z

|

⁢

e

i

⁢

θ

and

w

=

|

w

|

⁢

e

i

⁢

ϕ

with

|

ϕ

-

θ

|

<

π

. Then we have the following:

•

The phase of the geometric mean of

z

and

w

can be chosen to lie
between

θ

and

ϕ

. This is because, as described earlier, this
phase can be chosen as

(

θ

+

ϕ

)

/

2

.

•

The phase of the arithmetic mean of

z

and

w

can be chosen to lie
between

θ

and

ϕ

.

By a simple induction

argument

, these two facts imply that we can introduce
polar representations

a

n

=

|

a

n

|

⁢

e

i

⁢

θ

n

and

g

n

=

|

g

n

|

⁢

e

i

⁢

ϕ

n

where, for every

n

, we find that

θ

n

lies between

α

and

β

and likewise

ϕ

n

lies between

α

and

β

.
Furthermore, since

ϕ

n

+

1

=

(

ϕ

n

+

θ

n

)

/

2

and

θ

n

+

1

lies between

ϕ

n

and

θ

n

, it follows that

|

ϕ

n

+

1

-

θ

n

+

1

|

≤

1

2

⁢

|

ϕ

n

-

θ

n

|

.

Hence, we conclude that

|

ϕ

n

-

θ

n

|

→

0

as

n

→

∞

. By
the principle of nested intervals, we further conclude that the sequences

{

θ

n

}

n

=

0

∞

and

{

ϕ

n

}

n

=

0

∞

are both

convergent

and converge to the same limit.

Having shown that the phases converge, we now turn our attention to the
moduli. Define

m

n

=

max

⁡

(

|

a

n

|

,

|

g

n

|

)

. Given any two complex
numbers

z

,

w

, we have

|

z

⁢

w

|

≤

max

⁡

(

|

z

|

,

|

w

|

)

and

|

z

+

w

2

|

≤

max

⁡

(

|

z

|

,

|

w

|

)

,

so this sequence

{

m

n

}

n

=

0

∞

is

decreasing

. Since it

bounded from
below

by

0

, it converges.

Finally, we consider the ratios of the moduli of the

arithmetic

and geometric
means. Define

x

n

=

|

a

n

|

/

|

g

n

|

. As in the real case, we shall derive a
recursion relation for this quantity:

x

n

+

1

=

|

a

n

+

1

|

|

g

n

+

1

|

=

|

a

n

+

g

n

|

2

⁢

|

a

n

⁢

g

n

|

=

|

a

n

2

|

+

2

⁢

|

a

n

|

⁢

|

g

n

|

⁢

cos

⁡

(

θ

n

-

ϕ

n

)

+

|

g

n

|

2

2

⁢

|

a

n

⁢

g

n

|

=

1

2

⁢

|

a

n

|

|

g

n

|

+

2

⁢

cos

⁡

(

θ

n

-

ϕ

n

)

+

|

g

n

|

|

a

n

|

=

1

2

⁢

x

n

+

2

⁢

cos

⁡

(

θ

n

-

ϕ

n

)

+

1

x

n

For any real number

x

≥

1

, we have the following:

x

-

1

≥

0

(

x

-

1

)

2

≥

0

x

2

-

2

⁢

x

+

1

≥

0

x

2

+

1

≥

2

⁢

x

x

+

1

x

≥

2

If

0

<

x

<

1

, then

1

/

x

>

1

, so we can swithch the roles of

x

and

1

/

x

and
conclude that, for all real

x

>

0

, we have

x

+

1

x

≥

2

.

Applying this to the recursion we just derived and making use of the half-angle
identity for the cosine, we see that

x

n

+

1

≥

1

2

⁢

2

+

2

⁢

cos

⁡

(

θ

n

-

ϕ

n

)

=

cos

⁡

(

θ

n

-

ϕ

n

2

)

.

Title

complex arithmetic-geometric mean

Canonical

name

ComplexArithmeticgeometricMean

Date of creation

2013-03-22 17:10:05

Last modified on

2013-03-22 17:10:05

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

15

Author

rspuzio (6075)

Entry type

Result

Classification

msc 33E05

Classification

msc 26E60

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