Seeking a correct relationship between the number of zeros of an entire function and it's derivative.

388 Views Asked by At

As far as polynomials are concerned, we know by virtue of fundamental theorem of algebra that a polynomial of degree $n$ has exactly $n$ roots in $\mathbb{C}$ and its derivative then will have $n-1$ roots. Thus we can say that the number of roots of a polynomial of degree $n$ is greater than the number of roots of its derivative.

However, my concern is that can I get similar conclusion for transcendental entire functions ?

Note that for the entire function $e^z$ has no zeros in $\mathbb{C}$.

1

There are 1 best solutions below

6
On BEST ANSWER

That conclusion does not hold for transcendental entire functions. As an example, if $h$ is an entire function then $f(z) = e^{h(z)}$ has no zeros, but $f'(z) = h'(z)e^{h(z)}$ can have finitely many (take $h$ as a polynomial) or infinitely many (e.g. $h(z) = \sin(z)$) zeros.

There is also no general estimate in the other direction: $g(z) = 1 + e^z$ has infinitely many zeros, but $g'(z) = e^z$ has none.