proof of Rouché’s theorem
Consider the integral
where . By the hypotheses, the function is non-singular on or on the interior of and has no zeros on . Hence, by the argument principle, equals the number of zeros (counted with multiplicity) of contained inside . Note that this means that must be an integer.
for all on . By the triangle inequality, this implies that on . Hence is a continuous function of when and . Therefore, the integrand is a continuous function of and . Since is compact, it follows that is a continuous function of .
Now there is only one way for a continuous function of a real variable to assume only integer values – that function must be constant. In particular, this means that the number of zeros of inside is the same for all . Taking the extreme cases and , this means that and have the same number of zeros inside .
|Title||proof of Rouché’s theorem|
|Date of creation||2013-03-22 14:34:26|
|Last modified on||2013-03-22 14:34:26|
|Last modified by||rspuzio (6075)|