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The problem is to find the number of rotations around the origin for the function

$$f(z)=z^{2013}+2z+1 $$ when $z$ moves through $\left\{|z|=1\right\}$.

I tried to solve it with the help of argument principle.

$${{N}_{r}}\left( {{z}^{2013}}+2z+1 \right)={{N}_{\left| z \right|<1}}\left( {{z}^{2013}}+2z+1 \right),$$ where ${{N}_{r}}$ - number of rotations around origin, ${{N}_{\left| z \right|<1}}$- number of zeros of ${{z}^{2013}}+2z+1$ in $\left\{ z:\left| z \right|<1 \right\}$. As ${{z}^{2013}}+2z+1$ is continuous function and ${{z}^{2013}}+2z+1$has no roots on $\left\{ \left| z \right|=1 \right\}$, we can use the theorem of Rouché: if we take $r=0.999999$ then $$|z^{2013}+1|<r^{2013}+1<2r=|2z| ,$$ so, there is only 1 root in $\left\{ z:\left| z \right|<0.999999 \right\}$. But if we take $r=1.001$ and use this theorem again, $$|2z|<2r<||r|^{2013}-|1||\le|z^{2013}+1|,$$ so, in $\left\{|z|<1.001\right\}$ there are 2013 roots. But how many roots are there in $\left\{1<|z|<1.001\right\}$ and $\left\{0.999999<|z|<1\right\}$ -- is unknown. Anything usefull will be appreciated.

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2 Answers 2

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We note that $z=-1$ isn't a zero of $z^{2013}+2z+1$, therefore it doesn't vanish in some $D_{2\varepsilon}(-1)$. Consider the path $\gamma$ which tours the unit circle, except staying at distance $\varepsilon$ from $-1$. Along that curve, $|z^{2013}|\leq 1<|2z+1|$, hence $z^{2013}+2z+1$ has the same number of zeroes inside $\gamma$ (and therefore inside $S^1$) as $2z+1$, namely $1$.

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  • $\begingroup$ Great! It's that one step with which I could not come up. $\endgroup$
    – cool
    Commented Aug 6, 2013 at 13:29
  • $\begingroup$ I should have specified, you have to bypass $-1$ inside the unit disc, for the inequalities to hold. $\endgroup$ Commented Aug 6, 2013 at 14:45
  • $\begingroup$ It was obvious. $\endgroup$
    – cool
    Commented Aug 6, 2013 at 15:50
  • $\begingroup$ Yes, yes, but I think someone asked about it in a comment that's no longer there. $\endgroup$ Commented Aug 6, 2013 at 15:53
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If you don't only look at the statement of Rouché's theorem, but also at the proof, you will notice that the condition $\lvert f(z) - g(z)\rvert < \lvert g(z)\rvert$ on $\partial V$ is only used to guarantee that

$$g(z) + \lambda \bigl(f(z) - g(z)\bigr) = (1-\lambda)g(z) + \lambda f(z)$$

has no zeros on $\partial V$ for $0 \leqslant \lambda \leqslant 1$. That weaker condition is all that is needed to have the conclusion of the theorem.

Here, with $f(z) = z^{2013} + 2z + 1$ and $g(z) = 2z+1$, we have $\lvert f(z) - g(z) \rvert = \lvert z\rvert^{2013} = 1 \leqslant \lvert g(z)\rvert$ on $C = \{\lvert z\rvert = 1\}$, and equality holds only at $z = -1$, where we have $f(-1) = -2$ and $g(-1) = -1$, so $(1-\lambda)g(z) + \lambda f(z)$ has no zeros on $C$, and the conclusion of the theorem holds.

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