Characteristic of zeroes of a power series

Question

$f(z)= g_0+(g_1−1)z+g_2 z^{2}+g_3z^{3}...$ be a power series where all $g_i$'s are strictly greater than 0 and sum of all $g_i$'s is equal to 1 and $f'(1)>0$. I see in the book of J. Medhi it is written that exactly one real zero of the above series is in $(0,1)$ and modulus of other zeros of the series is strictly greater than 1. But I think that 1 is the zero of the series but I can't understand how all but one zeroes of the series is of modulus greater than 1. Please explain the statement "exactly one real zero of the above series is in (0,1) and modulus of other zeros of the series is strictly greater than 1". please help me. Thanking you in advance.

Answer 1

Note that $f(1)=0$. If there are two zeros, say $0<a<b<1$ in $(0,1)$ then there exists a point $\xi \in (a,b)$ such that $f'(\xi)=0$ and $\zeta \in (b,1)$ such that $f'(\zeta)=0$. This gives $g_1+2\xi g_2+3 g_3(\xi )^{2}+...=1=g_1+2\zeta g_2+3 g_3(\zeta )^{2}+...$. Since $\xi <b <\zeta$ this implies $g_2=g_3=... =0$, a contradiction. Hence there can be at most one zero of $f$ in $(0,1)$. To prove that there exists a zero in $(0,1)$ note that $f'(1)>0$ so $f(z) <f(1)=0$ for a value of $z$, say $z_0$ less than 1 but close to 1. Since $f$ is continuous and $f(0)=g_0>0$ it follows that $f$ must attain the value $0$ at some point between $0$ and $z_0$. We have proved that $f(1)=0$ and there exists exactly one point in $(0,1)$ where $f$ is $0$. Can you now show that there is no zero of $f$ in $(-\infty ,0)$? Once you do this it follows that any other zero of $f$ (meaning a zero other than 1 and the zero inside $(0,1$)) must be greater than 1.
To show that there are no complex zeros $z$ with $|z|<1$ other than the one in $(0,1)$ we need some complex analysis. Let $h(z)=z$ and note that $|f(z)+h(z)| <|h(z)|$ on the boundary of the unit disk. Rouche's Theorem show that $f$ has only one zero in $\{z:|z|<1\}$ which must be the one we obtained earlier.