0

Does the derivative of a holomorphic function $f(t)$ always vanish at multiple roots of $f$?

I know its true for polynomials, but is it a general fact known to hold for all transcendental functions?

gymbvghjkgkjkhgfkl
  • 8,090
crow
  • 1
  • Try expanding the power series around a multiple root. – Mark Schultz-Wu Mar 04 '17 at 04:55
  • $f$ is holomorphic, let $f(z)=(z-z_0)^mg(z)$ with $g(z_0)\neq0$ and $m>1$. – Nosrati Mar 04 '17 at 05:06
  • Ok.. ${\frac {\partial }{\partial t}} \left( \left( t-{\it t0} \right) ^{m} g \left( t \right) \right)={\frac { \left( t-{\it t0} \right) ^{m}mg \left( t \right) }{t-{\it t0}}}+ \left( t-{\it t0} \right) ^{m}{\frac {d}{dt}}g \left( t \right)$ – crow Mar 04 '17 at 05:16

1 Answers1

2

Yes. Suppose $f$ has a multiple root at $z_0$, so that $$\lim_{z\to z_0} \frac{f(z)}{(z-z_0)^2}$$ exists. Now let $g(z) = f(z)/(z-z_0)^2$, such that the value of $g(z_0)$ is given by the above limit. Then $g$ is holomorphic, and $$f'(z) = g'(z)(z-z_0)^2 + 2g(z)(z-z_0)$$ and so we see that $f'(z_0) = 0$.

florence
  • 12,819