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The question is:

by using rouche theorem, calculate number of zeros $F(z)=(z^2 + 2)(z^2 > + 1) - iz(z^2 + 2)$

where $D={\Im(z)> 0}$.

How do I need to choose $h(z)$ and $g(z)$ s.t $F= h + g$ so I can apply the theoremm and what about the condition $D={\Im(z)> 0}$?

I know the application of rouche theorem when we have condition on $|z|$, for example $1<|z|<2$ but I don't know what I need to do with the condition $\Im(z)> 0$.

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To tackle domains like this, try using the curve $\gamma_R$, which we define for $R>0$ to be the union of the upper half circle of radius $R$ with the segment $-R<x<R$ on the real line. Apply Rouche's Theorem with this curve, and check what happens as $R\rightarrow\infty$. – Robert Maschal Jun 2 '12 at 15:08
ok I got the idea. but then I could not divide F into two parts :S – Brhn Jun 2 '12 at 15:17
What is the meaning of $z^2>+1?$ for $z\in \mathbb{C}$? Is the polynomial $F(z)=(z^2+1)-iz(z^2+2)?$ – The Substitute Feb 9 '15 at 10:35

Homework problems tend to favor using $1$ as one of the function when applying Rouché's Theorem.

Try using $g(z) = \frac{F(z)}{(z^2+2)(z^2+1)}$, and $f(z) = 1$. Note that it is easy to see that $g$ has exactly two poles in the upper half plane ($i$, and i$\sqrt{2}$).

Now consider $|g(z)-1| = |\frac{z}{z^2+1}|$ on the curve suggested by @RobertMaschal above. Split the computation into two parts; the real axis and the semi-circle. It is straightforward to show that for $R$ sufficiently large, you will have $|g(z)-1| < 1$, hence by Rouché's Theorem, $g$ has the same number of poles and zeros in the region enclosed. Since $R$ is arbitrarily large, this means it is true on the upper half place. Hence $g$ has two zeros there, from which the result for $F$ follows.

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