$\displaystyle f(z) = z^5 + 15z +1$
Prove all roots of $\displaystyle f(z)$ are inside $\displaystyle |z| <1$
Question says use Rouche Theorem
Can someone help me get started on this?
see this link Rouché's theorem - Wikipedia, the free encyclopedia
First by the fundemental theorem of Algebra f has 5 zero's including multiplicity and Consider the Unit Disk.
Define $\displaystyle g(z)=z^5$ and note that g(z) has 5 zero's in the unit disk (i.e 0 is a root of multiplicity 5)
Now lets check Rouche's theorem $\displaystyle |f(z)-g(z)|< |f(z)|+|g(z)|$
$\displaystyle |f(z)-g(z)|=|15z+1| \le 15|z|+1 \le 16$ on the boundry of the unit disk
and $\displaystyle |f(z)|+|g(z)|=|z^5+15z+1|+|z^5|\ge 2|z^5|+15|z|+1 \ge 18$
so then
$\displaystyle |f(z)-g(z)|< |f(z)|+|g(z)|$ on the boundry so they gave the same number of zero's on the unit disk.
So f has 5 zero's inside the unit disk. Yay
There are different ways to write Rouche's Theorem. The simplest for me is:
If $\displaystyle f(z)$ and $\displaystyle g(z)$ are analytic inside and on a simple closed curve C and if $\displaystyle |g(z)|<|f(z)|$ on C, then f and f+g have the same number of zeros in C.
So let $\displaystyle f(z)=15z$ and $\displaystyle g(z)=z^5+1$ and C be the unit circle. Then $\displaystyle |f(z)|>|g(z)| $ on C so therefore $\displaystyle f(z)=15z $ and $\displaystyle f+g=z^5+15z+1$ have the same number of zeros inside C which is one.
Thanks. This looks correct shawsend
However am I correct in saying the following:
for all on circle
Because this will imply we have 5 zeros. In this I have used
$\displaystyle f(z)=z^5+15z+1$ and $\displaystyle g(z) = -15z-1$
(This is to bring the Rouche in-equality in the form you wrote ($\displaystyle |g(z)|<|f(z)|$)
If that is the case (which I think is) then there is some contradiction in what you have proved (that there is just one zero) and what this logic says (that all the 5 zeros are inside the unit circle).
Where am I messing up?
I don't think your inequality is valid. I used the code below to draw the zeros. It shows one zero in the unit circle (red dot are zeros).
Code:myEqn = z^5 + 15 z + 1; roots = N[z /. Solve[myEqn == 0, z]] points = Point@ {Re[#], Im[#]} & /@ roots Show[Graphics[{{PointSize[0.01], Red, points}, {Blue, Circle[{0, 0}, 1]}}, PlotRange -> {{-2, 2}, {-2, 2}}, Axes -> True]]