has a zero of order 5 at the origin, and that counts as five zeros for the purposes of Rouché's theorem. You must always count zeros according to their multiplicity.
Hey guys. I have the following theorem:
I'm supposed to count the zeros of, for example, z^5+15z+1 on D(0;2) (the open disk about 0 of radius 2). But this gives me contradictory results.Originally Posted by Rouche
Notice that |z^5+1|\geq ||z^5|-1|=31>30=|15z| on the contour . It follows by Rouche that and z^5+15z+1 have the same number of zeros on .
But now observe that |z^5|=32>31=15|z|+1\geq|15z+1| on the same contour So again by Rouche, and z^5+15z+1 have the same number of zeros on .
Putting these together, we see and have the same number of zeros on D(0;2). But we can see that has five zeros on D(0;2), while only has one zero.
Does anyone have a couple minutes to show me where I've gone wrong?
Thanks!
NOTE: The forum's latex compiler is apparently acting up, so forgive the code.