We note that
where
Now for |z| < 1,
and so we conclude that if
is violated, then 
.
Equality is possible in
only if z is real, and we easily check that for
,
only at 
.
For 
, equality holds in 
when
for 
; i.e., when
Let 
.
Then 
maps 
to the interior of the circle
one of whose diameters is
but 
maps 
to
so 
fails if 
.
Moreover, if
, 
still fails unless
We next prove that the only 
with |z| close to 
are negative real numbers.
Since 
intersects the closed set
only at 
, there exist 
, 
such that
fails for z in
It only remains to find the possible
elements of 
that lie in
For 
,
so if 
, then we must have
.
Since 
for 
,
and 
, to achieve
,
we must have 
for 
, say.
Then
where
Hence for |z| =r,
while
On the circle 
,
so
On the other hand, 
, so by Rouché's theorem 
and
have the same number of zeros inside
.
By the earlier part of the argument, and another application of Rouché's
theorem, 
and 
have the same number of
zeros inside 
, namely one.
Therefore 
has exactly one zero inside 
, and since 
has real coefficients, this zero has to be real. 
