Complex roots of $z^{10} - 6z^9 + 6^9$ inside the disk of radius six

Question

Show that there are exactly $8$ roots of the polynomial $z^{10} - 6z^9 + 6^9$ inside the disk of radius $6$ centered at the origin in the complex plane.

I am trying to use Rouche's theorem but am unable to find a suitable other function.

Answer 1

Note: Let us see the roots first and their modulus:

sage: R.<z> = PolynomialRing( QQ )
sage: f = z^10 - 6*z^9 + 6^9
sage: for root in f.roots( ring=QQbar, multiplicities=False ):
....:     print "modulus %s for root %s" % ( abs(CC(root)), root )
....:     
modulus 4.62158229715737 for root -4.367716331288758? - 1.510654486900259?*I
modulus 4.62158229715737 for root -4.367716331288758? + 1.510654486900259?*I
modulus 4.67794908447230 for root -2.530717800628705? - 3.934294733557240?*I
modulus 4.67794908447230 for root -2.530717800628705? + 3.934294733557240?*I
modulus 4.80918834167949 for root 0.4579587909670078? - 4.787333939837699?*I
modulus 4.80918834167949 for root 0.4579587909670078? + 4.787333939837699?*I
modulus 5.08528488001066 for root 3.518698291252366? - 3.671360053985808?*I
modulus 5.08528488001066 for root 3.518698291252366? + 3.671360053985808?*I
modulus 6.00409246947123 for root 5.921777049698090? - 0.9907991499946299?*I
modulus 6.00409246947123 for root 5.921777049698090? + 0.9907991499946299?*I

The problem is "hard to solve via Rouche's Theorem" because of the modulus of the last two roots, $$ a=x_9,\ \bar a=x_{10}\ , $$ (the two near $6\pm i$,) they are outside the circle, but they are very closed to the boundary. A perturbation of $f$ by using a suitable function $g$ such that using a continuous deformation of the coefficients from $f$ to $f+g$, for instance $t\to f+tg$, $t\in[0,1]$, which hopefully offers a better study for $f+g$,

is of course possible,

but showing "with bare hands" an inequality, constrained to use Rouche's Theorem, leads to a long search for a $g$ which also "almost vanishes" near $a$ (and $\bar a$).

The minimal value of $f$ on the boundary of the disc centered in $0$ with radius $6$ is (numerically) at most...

Sage code:

f = z^10 - 6*z^9 + 6^9
t = log( f.roots( ring=CC, multiplicities=0 )[-1] ).imag()
v = f( 6*exp(i*t) )
print ( "v = f( 6*exp(i*%s) ) is approximatively\nv ~ %s with\n|v| ~ %s"
        % ( t, v, abs(v) ) )

with results

v = f( 6*exp(i*0.165778932785284) ) is approximatively
v ~ 65236.8404263379 - 41086.7038660869*I with
|v| ~ 77097.0984109713
sage: 6^9
10077696

and the value $77097,09\dots$ is small compared to $6^9$. One conclusion is that it may be difficult to find a solution, constrained by the condition to use Rouche's Theorem.

Numerically we have a "true proof", by computing integrals around each solution numerically to given precision, e.g. (using the pari/gp calculator)

? f(z) = z^10 - 6*z^9 + 6^9;
? g(z) = 10*z^9 - 6*9*z^8;
? \p30
   realprecision = 38 significant digits (30 digits displayed)
? intnum( t = 0, 2*Pi, g( -13/3 + I*3/2 + exp(I*t) ) / f( -13/3 + I*3/2 + exp(I*t) ) * I * exp(I*t) ) / 2 / Pi
%17 = -1.53238199451739564033852920557 E-18 + 1.00000000000000000239209548711*I

and this isolates exactly one root of $f$ inside the disk centered in $-13/3+3i/2$ of radius one. In this century, this is a proof. (Similar proofs show that the Rieman zeta function has only predicted zeros in the critical strip up to "some human height".)

Similar integrals can locate the other roots.

---

My best try to solve the problem (humanly, but not using Rouche's Theorem) is as follows:

As in the comment of Lord Shark the Unknown, start with $f=0$, $$ z^{10}-6z^9 + 6^9=0\ ,$$ divide by $6^{10}$, and substitute $u=z/6$, the new equation is $$ u^{10}-u^9 + \frac 16=0\ ,$$ to be solved inside the unit circle.

Now multiply with $6$, divide by $u^{10}$ and substitute $w=1/z$, ($z=0$ is not a root,) to get the equation in $w$ $$ 6 - 6w + w^{10}=0\ ,$$ to be solve outside the unit circle. There should be eight solutions. Equivalently, we solve inside the unit circle, show, there are two solutions exactly. The situation is as follows:

Now show via analysis that for any $A\in (5,6]$ there is no solution of modulus one for the equation $$ w^{10} -6w+A=0\ .$$

(This is maybe connected with a study of the zeros of $\sin(10t) -6\sin(t)$.)

With this "cheat", we reach the limiting case $$ w^{10} -6w+5=0\ ,$$ which has the solution with multiplicity one $w=1$. By this deformation was to pass from the above distribution of zeros to the following one:

The progress is, that we have moved one the transcendental pair of complex conjugated roots "that hurts" first to a real "pair", the one for $A$ corresponding to $(w^{10}-6w+A)'=0$, i.e. $w_0=(6/10)^{1/9}$, $\displaystyle A_0=\left( 6-\frac 6{10}\right)\left( \frac 6{10}\right)^{1/9}$, then the pair splits in the real axis, till the the bigger one reaches the $1$.

If this "is O.K.", then we have to show that the quotient $$ \frac{ w^{10}-6w+A }{w-1} = w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w - 5 $$ has exactly one root in the unit disk. It is clear that there is exactly one real solution in the unit disk in the interval $(-1,1)$, monotony. We denote this solution by $b$. To have an idea about its magnitude,

sage: (w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w - 5 ).roots(ring=AA)
[(0.8794728676157584?, 1)]

so $0,8<b<0,9$. We divide the above with $(w-b)$, getting the equation $$ (1-b)w^8 +(1-b^2)w^7 +(1-b^3)w^6 +(1-b^4)w^5 +(1-b^5)w^4 +(1-b^6)w^3 +(1-b^7)w^2 +(1-b^8)w +(1-b^9) =0\ . $$ We have to show there are no solutions in the unit disk.

Here one "should be able" to apply Rouche's Theorem for passing from the above function corresponding to $b\approx 0.87947286761575\dots$ to the one corresponding to $$b^*=\frac 89\ .$$

The corresponding polynomial is now: $$ F=\frac {U(w)-U(8/9)}{w-8/9}\ , $$ where $$U=w^9 + w^8 + w^7 + w^6 + w^5 + w^4 + w^3 + w^2 + w\ .$$ This is $$F= w^{8} + \frac{17}{9} w^{7} + \frac{217}{81} w^{6} + \frac{2465}{729} w^{5} + \frac{26281}{6561} w^{4} + \frac{269297}{59049} w^{3} + \frac{2685817}{531441} w^{2} + \frac{26269505}{4782969} w + \frac{253202761}{43046721} \ .$$

Its roots are numerically

sage: F = w^8 + 17/9*w^7 + 217/81*w^6 + 2465/729*w^5 + 26281/6561*w^4 + 269297/59049*w^3 + 2685817/531441*w^2 + 26269505/4782969*w + 253202761/43046721
sage: Froots = F.roots(ring=QQbar, multiplicities=0)
sage: Froots
[-1.221780940364067? - 0.4246085183504945?*I,
 -1.221780940364067? + 0.4246085183504945?*I,
 -0.6862804721726701? - 1.078283076547882?*I,
 -0.6862804721726701? + 1.078283076547882?*I,
 0.1304408969654373? - 1.237226471865497?*I,
 0.1304408969654373? + 1.237226471865497?*I,
 0.8331760711268555? - 0.8344293526158634?*I,
 0.8331760711268555? + 0.8344293526158634?*I]

Now we consider the polynomial:

$$G= \left(w^2+\frac {22}9w+\frac 53\right) \left(w^2+\frac 43w+\frac 53\right) \left(w^2-\frac 29w+\frac {14}9\right) \left(w^2-\frac 53w+\frac 43\right) \ . $$ Its roots can be easily located. The difference $F-G$ is

sage: G = prod( [ w^2+a*w+b for a,b in ((22/9,5/3), (4/3,5/3), (-2/9,14/9), (-5/3,4/3)) ] )
sage: F-G
-1/27*w^6 + 83/729*w^5 + 3106/6561*w^4 + 29294/59049*w^3 + 274285/531441*w^2 + 2190635/4782969*w + 5196961/43046721

which is "small", and maybe can be controlled with $F$ or $G$. (Else use a better approximation.)

Have to stop here.