I used Rouche's theorem. I got $5$ solutions in $1<|z|<2$. Is my approach correct?
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1Is it? Can you give your details? – Angina Seng Oct 19 '18 at 19:34
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1Write your attempt in the question itself. – Mayuresh L Oct 19 '18 at 19:47
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For |z|<1, i choose g(z)=Z^2, f(z)=3z^5+1. Here f(z)>g(z). But for z|<1, g(z) should be dominating. So there is no solution here. For z|<2 I got 5 roots. Hence total 5 roots here. – Robin Oct 19 '18 at 19:52
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1You are very wrong to think that $|f|$ dominates $|g|$ on the boundary of the unit disc. (You only have to look at the boundary when applying Rouché's Theorem.) That is why you obtained a wrong conclusion. – Batominovski Oct 19 '18 at 20:10
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Can you explain in details? Thank you. – Robin Oct 19 '18 at 20:57
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All the solutions $z\in\mathbb{C}$ of $3z^5+z^2+1=0$ satisfy $|z|<1$. I am offering an elementary proof. You can also use Rouché's Theorem to prove this.
If $3z^5+z^2+1=0$, then $$3\,|z|^5=\left|3z^5\right|=\left|-z^2-1\right|\leq \left|-z^2\right|+|-1|=|z|^2+1\,.$$ If $|z|\geq 1$, then we have that $$3\,|z|^5>2\,|z|^5=|z|^5+|z|^5\geq |z|^2+1\,,$$ which is a contradiction. Thus, all the solutions $z\in\mathbb{C}$ of $3z^5+z^2+1=0$ satisfy $|z|<1$.
One can improve the bound by showing that the roots must satisfy $$0.7=\frac{7}{10}<|z|<\dfrac{\sqrt[3]{20}}{3}<0.905\,.$$ It is quite a delight to see that the upper bound is very sharp. The root with the maximum modulus has the modulus of around $0.9047$, and $\dfrac{\sqrt[3]{20}}{3}\approx 0.9048$. The lower bound is also not bad, as the root with the minimum modulus has the modulus of roughly $0.7208$.
Batominovski
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Use my solution as a hint of how to do it by Rouché's Theorem. This solution is a very strong hint. – Batominovski Oct 19 '18 at 19:59
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Note that for $|z|=1$, $$ |3z^5| = 3 > 1+1 \geq |z^2 + 1| $$ so Rouche implies there are 5 roots inside the unit disk. The main issue with your argument is that you need to check which function dominates on the boundary of the chosen contour.
Alvin Jin
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An illustrative plot
As $3z^5+z^2+1 = \phi(x,y) + i \psi(x,y) = 0$ we have $\phi(x,y)=0$ in red and $\psi(x,y) = 0$ in blue.
Cesareo
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