Different proof of fundamental theorem of algebra

I found this proof sketch in Rosenlicht's book. I get the overall idea, but i don't get why we can write $f(z)$ in the way shown in c). I would be grateful if someone explained this proof to me. Thank you.

• A similar proof appears in Rudin's PMA.
– lhf
Commented Dec 15, 2015 at 20:39

You can write $f(z)$ as a polynomial in $z-\zeta$. If you subtract off the constant term, which is $f(\zeta)$, you get a polynomial divisible by $(z-\zeta)$. If we let $m$ be the largest power of $(z-\zeta)$ that divides $f(z)-f(\zeta)$, then $\dfrac{f(z)-f(\zeta)}{(z-\zeta)^m}$ has a nonzero constant term, so let $a$ be this constant term and let $g(z)$ be the quotient of $\dfrac{f(z)-f(\zeta)}{a(z-\zeta)^m}-1$ by $(z-\zeta)$.
• How can I write $f(z)$ as a polynomial in $z-\zeta$? Thanks Commented Dec 15, 2015 at 20:06
• Write $w = z-\zeta$, and expand $g(w) := f(z) = f(w+\zeta)$. Commented Dec 15, 2015 at 20:18
• @DustanLevenstein Sorry, but how it will become a polynomial in $z-\zeta$, i.e. how will you get powers $(z-\zeta)^n$? Commented Dec 15, 2015 at 21:26
• @luka5z $g$ is a polynomial in $w$. When you put $z-\zeta$ in for $w$, you get a polynomial in $z-\zeta$. Commented Dec 15, 2015 at 22:04
• @DustanLevenstein I still don't get it how $f(\zeta)$ is a constant term of that polynomial... Commented Dec 16, 2015 at 9:55
If you show that all the complex derivatives of $f(z)$ exist (which makes $f(z)$ holomorphic), Taylor series can be used to derive this formula.
Such a formula can also be written for any holomorphic function. If the modulus of the holomorphic function has a minima (for example when its modulus is unbounded above large radial value $|z|$= $\sqrt{x^{2}+y^2}$ , z=x+iy ), then the minima must be zero.