For a &gt

Problem 11947

(American Mathematical Monthly, Vol.123, December 2016) Proposed by G. Stoica (Canada).

Letn be a positive integer, and let z1, . . . , zⁿ be the zeros in C ofzⁿ+ 1. For a > 0, prove 1

n

n

X

k=1

1

|zk− a|² = Pⁿ⁻1

k=0a^2k (1 + aⁿ)².

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Solution. For z ∈ C such that zⁿ6= −1, f (z) :=

Pⁿ⁻1 k=0z^2k

(1 + zⁿ)² = z²ⁿ− 1

(z²− 1)(1 + zⁿ)² = zⁿ− 1 (z²− 1)(1 + zⁿ). Let z^k= exp(πi(2k − 1)/n) for k = 1, . . . , n. If z^k6= −1 then

Res (f (z), zk) = Res

zⁿ−1 (z²−1)

1 + zⁿ, zk

!

= lim

z→z_k zⁿ−1 z2−1

nzⁿ⁻¹ = 2 n(z^k− z^k) On the other hand, if if n is odd then zⁿ= −1 and

z→−lim1(z + 1)²f (z) = lim

z→−1

 zⁿ− 1

z − 1 · z + 1 1 + zⁿ



= lim

z→−1

1 nzⁿ⁻¹ = 1

n,

z→−lim1(z + 1)



f (z) − 1 n(z + 1)²



= 0.

i) If n = 2m then z^k 6= −1, and by partial fraction decomposition, f (z) =

n

X

k=1

A^k

z − z^k where A^k= lim

z→z_k(z − z^k)f (z) = Res (f (z), z^k) = 2 n(z^k− z^k). Since zk= zn+1−k, it follows that Ak= −An+1−k and for a > 0

f (a) =

m

X

k=1

A^k a − zk

+

m

X

k=1

Aⁿ+1−k

a − zn+1−k

=

m

X

k=1

A^k

 1

a − zk

− 1

a − zk



=

m

X

k=1

A^k(z^k− z^k)

|z^k− a|² = 1 n

n

X

k=1

1

|z^k− a|². ii) If n = 2m + 1 then zk6= −1 for k 6= m + 1, and zm+1= −1.

Again by partial fraction decomposition, f (z) = 1

n(z + 1)² +

m

X

k=1

Ak

z − z^k +

n

X

k=m+2

Ak

z − z^k where A^k= 2 n(z^k− z^k). Since z^k= zⁿ+1−k, it follows that A^k= −Aⁿ+1−k and for a > 0

f (a) = 1 n(a + 1)² +

m

X

k=1

A^k a − z^k +

m

X

k=1

Aⁿ+1−k

a − zⁿ+1−k

= 1

n(a + 1)² +

m

X

k=1

A^k

 1

a − z^k − 1 a − z^k



= 1

n(−1 − a)² +

m

X

k=1

A^k(z^k− z^k)

|z^k− a|² = 1 n

n

X

k=1

1

|z^k− a|².

and we are done.