# Thread: Finding roots - Rouche's - calculating absolute value

1. ## Finding roots - Rouche's - calculating absolute value

I'm trying to find roots to a polynom using Rouche's theorem.

Rouche's
$\displaystyle \left| f(z) - g(z) \right| < \left| f(z) \right|-\left| g(z) \right|$

I'm feeling unsure of how to calculate the absolute values, if anyone could help...

I'll try an example, taken from wikipedia.

$z^5+3z+7$ let $f(z) = z^5$ and $g(z) = 3z + 7$

Examine |z| = 1
$|f(z)| = |z^5| = |z|^5 = |1|^5 = 1$

$|g(z)| = |3z+7| = |7 - (- 3z)| = |7|-|-3z| = 7 - 3z = 4$

$|f(z)| - |g(z)| = 1 - 4 = -3$

$|f(z) - g(z)| = |z^5 -3z -7| = |1 - 3 - 7| = |-9| = 9$

If I've done this correctly, can I deduce that there's no roots in |z| = 1? If not, how would I show that there's no roots in |z| = 1?

Please point out errors in my calculation. Furthermore, if I examine |z| = 2 I get |f(z)| - |g(z)| = 31 and |f(z) - g(z)| = 19 which indicates that there are 5 roots inside |z| = 2. Or..?

2. Let's see, I don't understand why do you use the difference, we have:

$F(z)=7+(z^5+3z)=f(z)+g(z)$ ( $f,g$ analytical functions on $|z|\leq 1$ )

On $|z|=1$ :

$|f(z)|=7$, $|g(z)|=|z^5+3z|\leq |z|^5+|3z|=4$

That is:

$|f(z)|>|g(z)|$ on $|z|=1$

According to Rouche's theorem, the number of roots of $F(z)$ on $|z|<1$ is equal to the number of roots of $f(z)$ on $|z|<1$ (including multiplicities).

As a consequence, $F(z)$ has no roots on $|z|<1$ .

Regards.

Fernando Revilla

3. Ok, thanks!

Did you assume no roots while setting f(z) g(z)?

$f(z) = z^5$ - has got 5 roots
$g(z) = 3z + 7$

$\left|f(z)\right| = \left|z^5\right| = \left|z\right|^5 = 2^5$
$\left|g(z)\right| = \left|3z+7\right| = \left|7-(-3z)\right| = 7-6 = 1$

So, |f(z)| > |g(z)| thus z⁵ + 3z + 7 has got 5 roots in |z| = 2. Or..?

4. Originally Posted by liquidFuzz
So, |f(z)| > |g(z)| thus z⁵ + 3z + 7 has got 5 roots in |z| = 2. Or..?
Right. On $|z|=2$ :

$|z^5|>|3z+7|$

So,

$F(z)=z^5+3z+7$ has 5 roots in $|z|<2$

Regards.

Fernando Revilla