Sketch of Second Proof (after Cauchy)
Consider the quantity s(z) = |P(z)|,
where P(z) = anz n + an -1z n -1 + ... + a1z1 + a0
is a polynomial of degree n. Clearly, s(z) 0.
If s(z) assumes a global minimum value, say, , on C, then
it suffices to prove that = 0, or
equivalently, it suffices to derive a contradiction from the alternative assumption that
> 0.
Suppose, then, that s(z0) = > 0.
It is convenient to "re-center" the arithmetic by setting Q(z) = P(z + z0 ), so |Q(0)| = |P(z0 )| = , where is likewise a global minimum for |Q(z)|.
Expanding out the definition of |Q(z)|, we obtain a set of new coefficients:
Q(z) = bnz n + bn -1z n -1 + ... + b1z1 + b0, where
bn = an 0, and
b0 = P(z0 ) 0.
Let m be the exponent of the lowest power of z in Q(z) whose coefficient bm is not zero. Now consider the behavior
of Q(z) for points z whose absolute value is very small, say the points
z = lying on a small circle
centered at the origin of radius . As observed
above, as z sweeps once around this small circle, Q(z) closely approximates
the behavior of bmzm + b0,
which sweeps out a small circle (of radius |bm| m)
around the point b0 = P(z0 ).
(In fact, Q(z) sweeps around this circle m times.) By choosing
sufficiently small, we may ensure that the
radius of this circle (that is, the magnitude of Q(z) - b0) is smaller
than . Such a circle will necessarily intersect
the line segment connecting the origin to the point b0 = P(z0 ),
at a point, say Q(z1 ), nearer the origin than b0 = P(z0 ).
But then |Q(z1 )| = |P(z1 + z0 )| < ,
contradicting our choice of as a global
minimum of s(z) = |P(z)|. (Actually, Q(z) may only lie near this circle, not on it. Fortunately, the discrepancy consists of the remaining terms of
Q(z) (if any), all of which contain powers of higher than m.
By taking even smaller, if necessary, we can
guarantee that this discrepancy does not wreck the geometry – see Figure 5.)
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Figure 5.Behavior of Q(z) for "small" values of z.
If z lies on the small circle of radius
centered at the origin of the z-plane, then
Q(z) bmzm + b0,
which is the larger circle in the figure, centered at Q(0) = b0,
with radius |bm| m.
For sufficiently small, the remaining terms of
(if any) of Q(z) sum to a vector whose magnitude (proportional to
n, n>m, is less than
the radius of the smaller circle in the figure.) As z sweeps out the small circle of
radius on the z-plane, the locus of
Q(z) loops around Q(0) = b0 (in fact, m times),
and necessarily intersects the vector drawn from the origin of the w-plane to
Q(0).
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Thus the assumption that > 0 is untenable, and we conclude that = 0,
that is, that P(z) has a complex root.
Copyright © 1996-2007 Alexander Bogomolny
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