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Find the number of solutions of $$z^4-7z^3-2z^2+z-3=0$$ inside the unit disc.

The Rouche theorem fails obviously. Is there any other method that can help?

I have known the answer by Matlab, but I have to prove it by complex analysis.

Thanks!

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  • $\begingroup$ Couldn't this be solved by algebra? $\endgroup$ Dec 28, 2014 at 15:39
  • $\begingroup$ Sorry, I'm preparing for a complex analysis exam. $\endgroup$ Dec 28, 2014 at 15:43
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    $\begingroup$ It should be $-2z^2$? $\endgroup$ Dec 28, 2014 at 15:43
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    $\begingroup$ This might be of help $\endgroup$
    – Varun Iyer
    Dec 28, 2014 at 15:45

2 Answers 2

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Let $f(z)=z^4-7z^3-2z^2+z-3$ and $g(z)=-7z^3$. Then, for $\lvert z\rvert=1$, $$ \lvert\, f(z)-g(z)\rvert=\lvert z^4-2z^2+z-3\rvert <7=\lvert -7z^3\rvert=\lvert g(z)\rvert. $$ Rouche Theorem provides that $f$ possesses exactly 3 roots inside the unit disc.

The only part to check is that $\lvert z^4-2z^2+z-3\rvert <7,$ which holds since we have the "$\le$" part, and the only way to have "$=$" is if every term in $z^4-2z^2+z-3$ is negative, which is impossible.

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Note that $|z^4-2z^2+z-3|<7$ if $|z|=1$. The triangle inequality only says "$\leq$" but equality can only occur if $z^4$, $-2z^2$, $z$ and $-3$ all have the same argument. So it suffices to check $z=-1$.

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