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What would be a good strategy to tackle this problem?

$$z^5 + {\sqrt2}z^3 + 1 = 0$$ where $z \in \Bbb C$.

This question comes out of my complex numbers text on finding roots of polynomials. I have tried to brute force this problem by factorising, however the $1$ hanging on the end leaves me confused.

Another approach I tried and couldn't work through any further was to express each complex variable in terms of the exponential function, $z=re^{i\theta}$ where $r$ is the modulus and $e^{i\theta}$ is our argument. Thus the above problem could be re-written as:

$$r^5e^{5i\theta} + \sqrt{2}r^3e^{3i\theta} + 1 = 0$$

I hope that helps give some context.

JJH
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    WolframAlpha has trouble finding exact solutions. That doesn't have to mean anything; it does not have a general formula for fifth-degree polynomial roots, so maybe it doesn't try very hard to factor the expression before just saying "Let's do numerical approximations instead". – Arthur Sep 23 '16 at 14:14
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    It depends. If you have any clue that it might have a solution that can be expressed in radicals, go for some factorization methods. If it's just an equation taken out of the blue, the chance is minimal that it does. Context is everything here and you haven't provided enough of it in your question. – Daniel R Sep 23 '16 at 14:29
  • @DanielR I made an edit above, I hope it helps give some further clarity. – JJH Sep 23 '16 at 15:02
  • @Authur I also tried WolframAlpha, and got the same response as you. :) – JJH Sep 23 '16 at 15:04
  • If you aren't sure as to whether the roots are in radical form or not, ensure it, or don't waste your time in this problem. – codetalker Sep 23 '16 at 15:32
  • you can find numerical values of the roots – Dr. Sonnhard Graubner Sep 23 '16 at 15:44
  • $\prod_{i = 1}^{5}r_{i} = -1$ where $r_{i}$ is a root. That gives some clue to find 'numerical approximations' to the roots. – Felix Marin Sep 23 '16 at 16:25

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$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

We can easily see that $$ \bbox[0.5em,#efe,border:0.1em groove navy]{\ r_{1} \equiv -0.78891677373356634396694435041559932834090342558\ } $$ is a 'good approximation' to the real root of $\ds{z^{5} + \root{2}z^{3} + 1 = 0}$.

Then, a 'fine approximation' is given by: \begin{align} z^{5} + \root{2}z^{3} + 1 & = \pars{z - r_{1}}\pars{z^{4} + bz^{3} + cz^{2} + dz - {1 \over r_{1}}} \\[5mm] & = z^{5} + \pars{b - r_{1}}z^{4}+ \pars{c - br_{1}}z^{3}+ \pars{d - cr_{1}}z^{2}+ \pars{dr_{1} - {1 \over r_{1}}}z + 1 \end{align} Now, we have an equation which can be solved analytically. Namely, $$ z^{4} + r_{1}z^{3} + \pars{r_{1}^{2} + \root{2}}z^{2} + r_{1}\pars{r_{1}^{2} + \root{2}}z - {1 \over r_{1}} = 0 $$
Felix Marin
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