Let D be the set of complex numbers having modulus less than 1 . For any fixed positive integer n and any distinct complex numbers z 1 ,...,z n in D , define

7.4.2. Let D be the set of complex numbers having modulus less than 1 . For any fixed positive integer n and any distinct complex numbers z 1 ,...,z n in D , define

W (z 1 ,...,z n )=

∑ 2 log |z

j −z k | + ∑ log |1 − z j z k |,

1 ≤ j<k≤n

j ,k=1

where z is the conjugate of the complex number z . (i) If n =2 , prove that the configuration (z 1 ,...,z n )∈D n that realizes the max-

imum of W is unique (up to a rotation) and it consists of two points that are symmetric with respect to the origin.

(ii) If n =3 , prove that the configuration that maximizes W is also unique (up to a rotation) and it consists of an equilateral triangle centered at the origin.

Vicent¸iu R˘adulescu Solution. (i) Take n = 2 and let z 1 ,z 2 be two distinct points in D. Then

W (z 1 ,z

2 ) = log(|z 1 | + |z 2 | − 2|z 1 | · |z 2 | · cos ϕ )

2 2 + log(1+ |z 2

1 | |z 2 | − 2|z 1 | · |z 2 | · cos ϕ

)+ log(1 − |z 2

1 | ) + log(1 − |z 1 | ), where ϕ denotes the angle between the vectors −→ Oz 1 and Oz −→ 2 . So, a necessary con-

dition for the maximum of W (z 1 ,z 2 ) is cos ϕ = −1, that is, the points z 1 , O, and z 2 are collinear, with O between z 1 and z 2 . Hence one may suppose that the points z 1 and z 2 lie on the real axis and −1 < z 2 <0<z 1 < 1. Define

f (z 1 ,z 2 ) = 2 log(z 1 −z 2 ) + 2 log(1 − z 1 z 2 ) + log(1 − z 2 1 ) + log(1 − z 2 2 ).

304 7 Inequalities and Extremum Problems Since the function log (1 − x 2 ) is concave on (0, +∞), it follows that

2 log

2 2 1 1 −z 2

(1 − z 1 ) + log(1 − z 2 ) ≤ 2log

2 On the other hand, it is obvious that 1 −z 1 z 2 ≤ 1 + (z 1 −z 2 ) 2 /4. Hence

which means that the maximum of f is achieved when z 1 = −z 2 . A straightforward computation shows that maxW = f (5 −1/4 , −5 −1/4 ) = 6 log 2 − 2 −1 · 5log5.

(ii) For n = 3, in order to maximize the functional W (z 1 ,z 2 ,z 3 ), it is enough to find the maximum of

F (z ,z ,z )=

1 2 3 ∏ |z j −z k | 3|z j −z k | + (1 − r j )(1 − r k ) 4· ∏ (1 − r j ),

1 ≤ j<k≤3 j =1

where r j = |z j |. Using the elementary identity

3 ∑ |z j | 2 = ∑ z j + ∑ |z j −z | 2 k ,

j =1

j =1

1 ≤ j<k≤3

we obtain

3 2 ∑ 2 |z j | ≥ ∑ |z j −z k | .

j =1

1 ≤ j<k≤3

Put S =∑ 3 j 2 =1 r j . We try to maximize F keeping S constant. Using the above inequal- ity, we have

2 ∏ ≤ j<k≤3 |z j −z k | ≤

1 |z j −z k |

3 ∑ j =1

|z j | 2 =S 3 , (7.7)

1 ≤ j<k≤3

∏ 2 (1 − r

2 2 2 ∏ 4 3|z

j −z k | + (1 − r j )(1 − r k )

1 ≤ j<k≤3

1 ≤ j<k≤3 |z j −z k | + (1 − r j )(1 − r k ) ≤ ⎣

7.5 Qualitative Results 305 ∑1 3 − ∑r 2

We have applied above the elementary inequality

1 ≤ j<k≤3

3 j =1

From (7.7), (7.8), and (7.9) we obtain

3 2 + 3S + 9 3

F ≤S 3 ·

(−S 4 + 27S) 3 .

It follows that the maximum of F (so the maximum of W ) is achieved if S = 3 ·4 −1/3 and max F =3 6 ·4 −4 , with equality when we have equality in (7.7), (7.8), and (7.9), that is, if and only if z 2 = ε z 1 ,z 3 = ε 2 z 1 , where ε = cos(2 π /3) + i sin(2 π /3). This implies that maxW = 6 log 3 − 8log2. ⊓ ⊔

Open problems. (i) Find the configuration that maximizes W (z 1 ,...,z n ), pro- vided that n ≥ 4. Is this configuration given by a regular n-gon, as for n = 2 or n = 3?

(ii) Study whether the maximal configuration “goes to the boundary” as n →∞, in the following sense: for given n, let z = (z 1 ,...,z n ) be an arbitrary configuration that realizes the maximum of W (z 1 ,...,z n ) and set a n = min{|z j |; 1 ≤ j ≤ n}. Is it true that a n →1 as n→∞?