Prove all roots of are inside
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 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
on the boundry of the unit disk
and
so then
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 and are analytic inside and on a simple closed curve C and if on C, then f and f+g have the same number of zeros in C.
So let and and C be the unit circle. Then on C so therefore and 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
and
(This is to bring the Rouche in-equality in the form you wrote ( )
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]]