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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=0a2k (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=0z2k

(1 + zn)2 = z2n− 1

(z2− 1)(1 + zn)2 = zn− 1 (z2− 1)(1 + zn). Let zk= exp(πi(2k − 1)/n) for k = 1, . . . , n. If zk6= −1 then

Res (f (z), zk) = Res

zn−1 (z2−1)

1 + zn, zk

!

= lim

z→zk zn1 z21

nzn−1 = 2 n(zk− zk) 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 zk 6= −1, and by partial fraction decomposition, f (z) =

n

X

k=1

Ak

z − zk where Ak= lim

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

f (a) =

m

X

k=1

Ak a − zk

+

m

X

k=1

An+1−k

a − zn+1−k

=

m

X

k=1

Ak

 1

a − zk

− 1

a − zk



=

m

X

k=1

Ak(zk− zk)

|zk− a|2 = 1 n

n

X

k=1

1

|zk− 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 − zk +

n

X

k=m+2

Ak

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

f (a) = 1 n(a + 1)2 +

m

X

k=1

Ak a − zk +

m

X

k=1

An+1−k

a − zn+1−k

= 1

n(a + 1)2 +

m

X

k=1

Ak

 1

a − zk − 1 a − zk



= 1

n(−1 − a)2 +

m

X

k=1

Ak(zk− zk)

|zk− a|2 = 1 n

n

X

k=1

1

|zk− a|2.

and we are done. 

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