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The Q is following : Prove that for each w in the unit disc $D(0, 1)$, the equation $z^5 (z − 2) = w $ has exactly five solutions in the unit disc counted with multiplicity.

My Approach : let $f(z) = z^5 (z − 2) - w $ Now $w \in D(0, 1)$. I apply Rouche's Theorem on $f$ on the Disk $D(0,1)$ and thus I get $g(z) = 2z^5$ which has $5$ roots in the given Disk and hence proved.

Is this approach correct ?

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  • $\begingroup$ Are we doing your homework? $\endgroup$
    – Matthew Levy
    Mar 29, 2015 at 5:03
  • $\begingroup$ I am trying out problems. If you wish, you could assume so... I will delete the Q. $\endgroup$
    – Rusty
    Mar 29, 2015 at 5:05
  • $\begingroup$ I certainly wouldn't delete it, it seems to me a perfectly reasonable question. $\endgroup$
    – pjs36
    Mar 29, 2015 at 5:11
  • $\begingroup$ Is my approach correct ? $\endgroup$
    – Rusty
    Mar 29, 2015 at 5:41
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    $\begingroup$ Yes, your approach is correct. Why are you asking? $\endgroup$
    – Alexandre Eremenko
    Mar 29, 2015 at 14:18

1 Answer 1

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Yes. The proof is correct. $ \:\: $

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