Advanced book on Mathematics Olympiad

402 Algebra

z+z 1

z−z 1

+

z+z 2

z−z 2

+···+

z+zn

z−zn

have negative real part, while ifzhas absolute value greater than 1, all terms in this sum
have positive real part. In order for this sum to be equal to zero,zmust have absolute
value 1. This completes the proof.
An alternative approach to this last step was suggested by R. Stong. Taking the real
part of

z+z 1

z−z 1

+

z+z 2

z−z 2

+···+

z+zn

z−zn

= 0 ,

we obtain

∑n

j= 1

Re

(

z+zj

z−zj

)

=

∑n

j= 1

1

|z−zj|^2

Re((z+zj)(z ̄− ̄zj))=

|z|^2 −|zj|^2

|z−zj|^2

.

Since|zj|=1 for allj, we conclude that|z|=1.

Remark.Whena=−i,φais called the Cayley transform.

178.Let the zeros of the polynomial bep, q, r, s. We havep+q+r+s=0,pq+pr+
rs+qr+qs+rs=−2, and hencep^2 +q^2 +r^2 +s^2 = 02 − 2 (− 2 )=4. By the Cauchy–
Schwarz inequality,( 1 + 1 + 1 )(q^2 +r^2 +s^2 )≥(q+r+s)^2. Furthermore, becauseq,r,s
must be distinct, the inequality is strict. Thus 4=p^2 +q^2 +r^2 +s^2 >p^2 +(−p)

2

3 =

4 p^2
3 ,
or|p|<

√

3. The same argument holds for the other zeros.
   (Hungarian Mathematical Olympiad, 1999)

179.We argue by induction onk. Fork=1 the property is obviously true.
Assume that the property is true for polynomials of degreek−1 and let us prove it
for the polynomialsPn(z),n≥1, andP(z)of degreek. Subtracting a constant from
all polynomials, we may assume thatP( 0 )=0. Order the zeros ofPn(z)such that
|z 1 (n)|≤|z 2 (n)|≤···≤|zk(n)|. The productz 1 (n)z 2 (n)···zk(n), being the free term
ofPn(z), converges to 0. This can happen only ifz 1 (n)→0. So we have proved the
property for one of the zeros.
In general, the polynomial obtained by dividing a monic polynomialQ(z)byz−
adepends continuously onaand on the coefficients ofQ(z). This means that the
coefficients ofPn(z)/(z−z 1 (n))converge to the coefficients ofP (z)/z, so we can apply
the induction hypothesis to these polynomials. The conclusion follows.

Remark.A stronger result is true, namely that if the coefficients of a monic polynomial
are continuous functions of a parametert, then the zeros are also continuous functions
oft.