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I am struggling to understand the solution below. I understand how to apply Rouches theorem when showing that there are a certain number of zeroes in a circle / annulus.

In this example (where they ask to show that there is a zero in each quadrant, I am not so sure. I dont really understand why they use inequalities with the real and imaginary parts.

I also dont understand why you can say that since since each of the roots of $z^4+1$ lie in a separate quadrant, then they must lie in separate quadrants of $z^4+z+1$ also.

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Rouché's theorem says that if $f$ and $g$ are analytic inside and on a closed contour $\Gamma$ and $|g| < |f|$ on $\Gamma$, then $f$ and $f+g$ have the same number of zeros inside $\Gamma$.

The inequalities are showing that $|g| < |f|$ on the pieces of axes and circle that make up the boundary of each pie-slice. Once you have that, and you know that $f$ has one root inside each pie-slice, Rouché tells you that $f+g$ has the same property.

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  • $\begingroup$ ok thank you. I still dont understand how each of the inequalities are made. For instance, 0<R, where does it come from? $\endgroup$ – Archie Judd Aug 17 '14 at 18:15
  • $\begingroup$ @ArchieJudd at first a fact about reals is given which later used in the first two cases where R=|x| and R=|y| respectively. $\endgroup$ – next Aug 17 '14 at 19:26
  • $\begingroup$ There is also a standard way to do such problems using the argument principle instead of Rouche's theorem. $\endgroup$ – Libertron Aug 17 '14 at 19:37
  • $\begingroup$ shouldn't you have to consider each separate pie-slice? i.e. for the first quadrant Re(z)>0 and Im(z)>0 etc. $\endgroup$ – Archie Judd Aug 18 '14 at 12:10
  • $\begingroup$ @ArchieJudd Note that $|f(z)|$ and $|g(z)|$ are invariant under reflection in the coordinate axes. So if it works for one slice, it works for the other three as well. $\endgroup$ – Robert Israel Aug 18 '14 at 18:05

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