# Math Help - Question on Rouche Theorem -1

1. ## Question on Rouche Theorem -1

$f(z) = z^5 + 15z +1$

Prove all roots of $f(z)$ are inside $|z| <1$

Question says use Rouche Theorem

Can someone help me get started on this?

2. Originally Posted by aman_cc
$f(z) = z^5 + 15z +1$

Prove all roots of $f(z)$ are inside $|z| <1$

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 $g(z)=z^5$ 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 $|f(z)-g(z)|< |f(z)|+|g(z)|$

$|f(z)-g(z)|=|15z+1| \le 15|z|+1 \le 16$ on the boundry of the unit disk

and $|f(z)|+|g(z)|=|z^5+15z+1|+|z^5|\ge 2|z^5|+15|z|+1 \ge 18$

so then

$|f(z)-g(z)|< |f(z)|+|g(z)|$ 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

3. Originally Posted by TheEmptySet
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 $g(z)=z^5$ 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 $|f(z)-g(z)|< |f(z)|+|g(z)|$

$|f(z)-g(z)|=|15z+1| \le 15|z|+1 \le 16$ on the boundry of the unit disk

and $|f(z)|+|g(z)|=|z^5+15z+1|+|z^5|\ge 2|z^5|+15|z|+1 \ge 18$

so then

$|f(z)-g(z)|< |f(z)|+|g(z)|$ 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

If we put z = -1 aren't we getting 16? I think the inequality you have used is wrong

Also, Rouche Theorem in my book says
$|f(z) - g(z)| < |f(z)|$

and not
$|f(z) - g(z)| < |f(z)| +|g(z)|$

4. If this is correct and someone can help me to the proof I guess then we would have solved it

$|15z + 1| < |z^5+15z + 1|$ for all $z$ on circle $|z|=1$

5. There are different ways to write Rouche's Theorem. The simplest for me is:

If $f(z)$ and $g(z)$ are analytic inside and on a simple closed curve C and if $|g(z)|<|f(z)|$ on C, then f and f+g have the same number of zeros in C.

So let $f(z)=15z$ and $g(z)=z^5+1$ and C be the unit circle. Then $|f(z)|>|g(z)|$ on C so therefore $f(z)=15z$ and $f+g=z^5+15z+1$ have the same number of zeros inside C which is one.

6. Originally Posted by shawsend
There are different ways to write Rouche's Theorem. The simplest for me is:

If $f(z)$ and $g(z)$ are analytic inside and on a simple closed curve C and if $|g(z)|<|f(z)|$ on C, then f and f+g have the same number of zeros in C.

So let $f(z)=15z$ and $g(z)=z^5+1$ and C be the unit circle. Then $|f(z)|>|g(z)|$ on C so therefore $f(z)=z$ and $f+g=z^5+15z+1$ 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
$f(z)=z^5+15z+1$ and $g(z) = -15z-1$
(This is to bring the Rouche in-equality in the form you wrote ( $|g(z)|<|f(z)|$)

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?

7. 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]]

8. Thanks very much shawsend !!!