0
$\begingroup$

If $f$ is a rational function defined on the complex plane. Then the number of the zeros is equal to the number of the poles (counting multiplicity) and considering points at infinity.

I can imagine a proof using the argument principle. Is this the simplest route to the theorem above? Is there a name for the theorem above?

  • 3
    Isn't this clear just from factoring the numerator and denominator then cancelling any common factors? The number of zeros and poles is then visible (being careful about those at infinity)2011-10-09
  • 0
    Just so I can clarify what your saying: $f(z)=\frac{(z-1)}{(z-2)(z-3)}$ has a zero at 1 and a pole at 2,3 each of multiplicity 1. How can I work out the multiplicity of the zero at infinity without invoking the theorem mentioned above?2011-10-09
  • 2
    A factor $z-\alpha$ has a zero of order $1$ at $\alpha$ and a pole of order $1$ at $\infty$.2011-10-09
  • 0
    Ok thanks guys. Really helped.2011-10-09

1 Answers 1