A Nice Factorable Quintic | Problem 215

▶ Greetings, everyone! Welcome to @aplusbi 🧡🤩💗
This channel is dedicated to the fascinating realm of Complex Numbers. I trust you'll find the content I'm about to share quite enjoyable. My initial plan is to kick things off with informative lectures on Complex Numbers, followed by a diverse range of problem-solving videos.

When you purchase something from here, I will make a small percentage of commission that helps me continue making videos for you. ❤️ ❤️ ❤️

❤️ This is Problem 215 on this channel!!! ❤️
🤩 Don't forget to SUBSCRIBE, hit that NOTIFICATION bell and stay tuned for upcoming videos!!!

▶ The world of Complex Numbers is truly captivating, and I hope you share the same enthusiasm! Come along with me as we embark on this exploration of Complex Numbers. Feel free to share your thoughts on the channel and the videos at any time.
▶ MY CHANNELS
Future channels: TBD
▶ EQUIPMENT and SOFTWARE
Camera: none
Microphone: Blue Yeti USB Microphone
Device: iPad and apple pencil
Apps and Web Tools: Notability, Google Docs, Canva, Desmos
LINKS

#complexnumbers #aplusbi #jeeadvanced #jee #complexanalysis #complex #jeemains
via @YouTube @Apple @Desmos @GoogleDocs @canva @NotabilityApp @geogebra

[NadiehFan]

To factor the left hand side of the equation

z⁵ + z − i = 0

we can start by substituting z = iw with gives

iw⁵ + iw − i = 0

and dividing both sides by i this gives

w⁵ + w − 1 = 0

so we got rid of the nonreal coefficient i. Now, since 5 ≡ 2 (mod 3) it is easy to see that the complex cube roots of −1 are zeros of w⁵ + w − 1 because if ζ is a complex cube root of −1 then ζ⁵ + ζ − 1 = ζ³·ζ² + ζ − 1 = −ζ² + ζ − 1 = −(ζ² − ζ + 1) = 0 since ζ³ + 1 = (ζ + 1)(ζ² − ζ + 1) implies that ζ² − ζ + 1 = 0 if ζ is a complex cube root of −1.

So, since both of the zeros of w² − w + 1 are zeros of w⁵ + w − 1 it follows that w² − w + 1 is a factor of w⁵ + w − 1 and by doing a polynomial division we find that we can write the equation w⁵ + w − 1 = 0 as

(w² − w + 1)(w³ + w² − 1) = 0

So, two of the roots of this equation are w₁ = ½ + ½i√3 and w₂ = ½ − ½i√3 and since z = iw this means that two of the roots of the original equation in z are z₁ = −½√3 + ½i and z₂ = ½√3 + ½i.

The other three roots of w⁵ + w − 1 = 0 are the roots of the cubic equation w³ + w² − 1 = 0. This cubic has a single real root and two complex conjugate roots, so z⁵ + z − i = 0 also has a single purely imaginary solution, but the expressions for these three roots are not nice.

Since z = iw implies w = −iz we can get a factorization of our original equation in z by substituting w = −iz in the factored form of our quintic equation in w which gives

(−z² + i z + 1)(iz³ − z² − 1) = 0

and by multiplying the first factor by −1 and the second factor by −i (and therefore also the right hand side by i) this can be written as

(z² − iz − 1)(z³ + iz² + i) = 0

which is of course the factorization of the original equation found at 7:09 in the video.

[NadiehFan]

Inspired by your comment that there is no quintic formula (which is _inaccurate_ since there _are_ formulas to solve quintic equations even though the roots of a general quintic cannot be expressed by radicals in terms of its coefficients) and inspired by your suggestion that you can try to factor a quintic I would suggest to try to solve this quintic equation:

z⁵ + 5z³ + 5z − 1 = 0

Using the rational root theorem or trying to factor this will get you nowhere, so don't waste your time on that. But it _is_ solvable in at least two different ways. Can you do it?

[error__brain_deleted]

Fixed-point iteration x←(1–x)^(1/5), starting with x=0.5, yields x=0.7548777... Thus, z=xi=0.7548777i
The other four roots I'll leave as an exercise.

[Blaqjaqshellaq]

4:08 Shouldn't the coefficient of z^3 be (b-a^2-i/c)?

[ThAlEdison]

when I ran it though Italian guy's formula, I got
but it's fairly simple to manipulate it to get it to match.