Algebra 401
An exercise in the section on induction shows that for any positive integerk,|sinkt|≤
k|sint|. Then
rn=
sint
sin(n− 1 )t
≥
1
n− 1
.
This implies the desired inequality|z|=r≥ n
√
1
n− 1.
(Romanian Mathematical Olympiad, proposed by I. Chi ̧tescu)
176.By the theorem of Lucas, if the zeros of a polynomial lie in a closed convex domain,
then the zeros of the derivative lie in the same domain. In our problem, change the
variable toz=^1 xto obtain the polynomialQ(z)=zn+zn−^1 +a. If all the zeros of
axn+x+1 were outside of the circle of radius 2 centered at the origin, then the zeros
ofQ(z)would lie in the interior of the circle of radius^12. Applying the theorem of Lucas
to the convex hull of these zeros, we deduce that the same would be true for the zeros of
the derivative. ButQ′(z)=nzn−^1 +(n− 1 )zn−^2 hasz=n−n^1 ≥^12 as one of its zeros,
which is a contradiction. This implies that the initial polynomial has a root of absolute
value less than or equal to 2.
177.The problem amounts to showing that the zeros ofQ(z)=zP′(z)−n 2 P(z)lie on
the unit circle. Let the zeros ofP(z)bez 1 ,z 2 ,...,zn, and letzbe a zero ofQ(z). The
relationQ(z)=0 translates into
z
z−z 1
+
z
z−z 2
+···+
z
z−zn
=
n
2
,
or
(
2 z
z−z 1
− 1
)
+
(
2 z
z−z 2
− 1
)
+···+
(
2 z
z−zn
− 1
)
= 0 ,
and finally
z+z 1
z−z 1
+
z+z 2
z−z 2
+···+
z+zn
z−zn
= 0.
The terms of this sum should remind us of a fundamental transformation of the complex
plane. This transformation is defined as follows: foraa complex number of absolute
value 1, we letφa(z)=(z+a)/(z−a). The mapφahas the important property that it
maps the unit circle to the imaginary axis, the interior of the unit disk to the half-plane
Rez<0, and the exterior of the unit disk to the half-plane Rez>0. Indeed, since
the unit disk is invariant under rotation by the argument ofa, it suffices to check this
fora=1. Thenφ(eiθ)=−icotθ 2 , which proves that the unit circle maps to the entire
imaginary axis. The map is one-to-one, so the interior of the unit disk is mapped to that
half-plane where the origin goes, namely to Rez<0, and the exterior is mapped to the
other half-plane. Ifzhas absolute value less than one, then all terms of the sum