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let $f(z)=z^9+z^5+8z^3+2z+1$, well, $$|f(z)-z^9|=|z^5+8z^3+2z+1|<|2|^9\text{ at } |z|=2$$ so $f$ has $9$ zeroes inside $|z|<2$

and $$|f(z)-8z^3|=|z^9+z^5+2z+1|<5<8|z|^3\text{ at } |z|=1$$ so inside $|z|<1$ $f$ has 3 zeroes, so $6$ zeroes are there inside $1<z<2$, could any one tell me whether I have properly applied Rouché’s theorem to this problem? Thank you for reply.

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ah! I am sorry I could not properly hyperlinked the page – Un Chien Andalou Jul 22 '12 at 20:17
Fixed it. URIs involving accents and stuff often don't work the way they should. – Dylan Moreland Jul 22 '12 at 20:27
You are correct and have supplied a complete proof. Source: I just finished a course in Complex Analysis at my university. – pre-kidney Jul 22 '12 at 20:32
thank you all,,, – Un Chien Andalou Jul 22 '12 at 21:26
up vote 3 down vote accepted

MSE community hereby certifies the proof as correct and complete. (Maybe add a word about no roots on boundary circles).

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Thank you dear sir – Un Chien Andalou Jul 23 '12 at 21:01

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