Real roots of complex polynomials

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
fraggle
Messages
18
Reaction score
0

Homework Statement



Let f be a polynomial of degree n >= 1 with all roots of multiplicity 1 and real on R. Prove that
f has at most one more real root than f'
f' has no more nonreal roots than f

Homework Equations



We are given the Gauss Lucas theorem: Every root of f' is contained in the convex hull of the roots of f.
Also previously proved is that if zk is a root of multiplicity 1 of f(z) then f'(zk) !=0 (zk is not a root of the derivative of f)

we express f(z)=c(z-z1)...(z-z2) where for each k f(zk)=0.

The Attempt at a Solution



I'm lost at where to begin, I've tried looking at it different ways but am not seeing where the difference of real and non real roots comes in.
Any suggestions? (if you can help I'd prefer hints rather than the whole answer)

Thanks
 
Physics news on Phys.org
R is a line in C, what can you say about the convex hull of collinear points?