# How many roots does $g(z)=z^7-2z^5+6z^3-z+1$ have inside the unit disk - Rouche's Theorem Application Verification

$$g(z)=z^7-2z^5+6z^3-z+1$$

and choose $$f(z)=2z^5-6z^3$$.

On $$\mid z\mid =1$$, we have

$$\mid f(z)-g(z) \mid=1$$ and $$\mid f(z) \mid=4$$

so $$\mid f(z)-g(z)\mid \leq \mid f(z)\mid$$

and by Rouche's Theorem, $$f(z)$$ and $$g(z)$$ have the same number of zeros inside the unit disk.

Now, $$f(z)=2z^5-6z^3=2z^3(z^2-3)$$ has 3 roots inside the unit disk, so $$g(z)$$ has 3 roots inside the unit disk.

Is this argument correct?

• why is $|f(z)|=4$ on $|z|=1$? Particularly, $f(i) = 8i$ Dec 18, 2018 at 22:07

I'm not sure about that. $$f(z)-g(z)=z^7+4z^5-12z^3+z-1$$. Then $$\lvert f(z)-g(z)\rvert\ne1$$ for some $$z$$ along the unit circle. Even if you wanted $$f(z)+g(z)=z^7-z+1$$, choosing $$z=i$$, we get $$i^7-i+1=-2i+1$$ which has modulus greater than $$1$$. So, this doesn't work either.

Rather, try comparing the coefficients. We can see that the largest coefficient is $$6$$, so along the unit circle the term $$6z^3$$ probably dominates the others. Take $$f(z)=6z^3$$ and $$h(z)=z^7-2z^5-z+1$$. Then $$\lvert h(z)\rvert\le \lvert z\rvert^7+2\lvert z\rvert^5+\lvert z\rvert +1\le 5,$$ $$\lvert f(z)\rvert= \lvert 6z^3\rvert=6.$$ So, $$g(z)$$ has as many zeros on the unit disk as $$f(z)=6z^3$$. $$f(z)$$ has a zero of multiplicity $$3$$ at $$z=0$$, so we conclude that $$g$$ has $$3$$ zeros in the unit disk.