Question: Find the number of zeroes of the equation [tex]$ z^7 - 2z^5 + 6z^3 - z+1 = 0 $[/tex], in the unit disk. We have Rouche's theorem which says that if f(z) and g(z) are two functions analytic in a neighborhood of the closed unit disk and if |f(z) - g(z)| < |f(z)| for all z on the boundary of the disk, then f and g have the same number of zeroes in the disk. Now there are easier problems which I can do just fine. But here is what I try. Now first off I know from here-say and mathematica that there are three roots inside the unit disk. So naturally I could use g(z) is either "6z^3" or "-2z^5 + 6z^3" since both have three zeroes (counting multiplicity) INSIDE the disk. So here is all i can see: using the identity [tex]$\right|a+b\right| \le \left|a\right|-\left|b\right|$ [/tex]over and over again I get: |f(z)| >= 1 immediately not very helpful.... from below I can see, using [tex]$g(z) = -2z^5 + 6z^3$[/tex], [tex]$\left|f(z)-g(z)\right| = \left|z^7 - z + 1\right| < 3$ [/tex] where i deliberately use strict < if you think about the geometry of this triangle inequality application.... ok so I cant get my |f(z)-g(z)| to be smaller than |f(z)|.... any help?