I was working on a problem from Gamelin; where I was required to find out zeros of $2z^5+6z^1-1$ , in the unit disk (in $\mathbb C$). I applied Rouché's theorem and find out zeros in the unit disk and I got to know that there is only one zero inside it.
Further, I have to show that it has one zero inside $(0,1)$ : Following is my answer and I am not sure about it. please correct me if i am wrong.
- This polynomial have only one zero inside the open unit disk. Therefore the only root that exist in unit disk must be a real one. Because complex roots exist only in pairs.
- Now we have to show that this zero is positive. Is it right if I say that, there is 1 changes in sign of the function's coefficients, so there will be at most 1 positive roots (maybe less). And now put -z on the place of z then all coefficients of the polynomial will turn negative. Therefore there is no negative root. So the only root that we have in the unit disk must be real; so it will lie in $(0,1)$.
Hint: $2\cdot 0^5 + 6\cdot 0-1 < 0 < 2\cdot 1^5+6\cdot 1-1$.