A Special Quadratic Equation | Problem 479

▶ 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.
FIRST PROBLEM: • Simplifying An Express... ❤️❤️❤️
Old But Gold: • Solving for a Complex ...
🤩🤩🤩 A Very Interesting Exponential Formula: zᵃ⁺ᵇⁱ = c+di: • A Very Interesting Exp...
🤩🤩🤩 An Interesting Locus Problem | Problem 399: • An Interesting Locus P...
❤️ ❤️ ❤️ My Amazon Store: www.amazon.com...
When you purchase something from here, I will make a small percentage of commission that helps me continue making videos for you. ❤️ ❤️ ❤️
Recently updated to display "Books to Prepare for Math Olympiads" Check it out!!!
❤️ An Exponent That Doubles | Problem 353: • An Exponent That Doubl...
🤩 Playlist For Lecture videos: • Lecture Videos
🤩 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
Main channel: / @sybermath
Shorts channel: / @shortsofsyber
This channel: / @aplusbi
Future channels: TBD
▶ Twitter: x.com/SyberMath
▶ 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
en.wikipedia.o...
/ @sybermath
/ @shortsofsyber
#complexnumbers #aplusbi #jeeadvanced #jee #complexanalysis #complex #jeemains
via @UA-cam @Apple @Desmos @GoogleDocs @canva @NotabilityApp @geogebra

[@NadiehFan]

At 8:58 you have
(2b² + 2b + k)² = (4k − 5)b² + (4k − 5)b + (k² − 1)
Take a good look at the coefficients of the quadratic in b at the right hand side. Both the coefficient of b² _and_ the coefficient of b are zero if 4k − 5 = 0, that is, if k = ⁵⁄₄. With this value of k we have k² − 1 = (⁵⁄₄)² − 1 = ²⁵⁄₁₆ − ¹⁶⁄₁₆ = ⁹⁄₁₆ = (³⁄₄)² and the equation reduces to
(2b² + 2b + ⁵⁄₄)² = (³⁄₄)²
which is easy to solve and which has b = −¹⁄₂ as its only real solution.
However, you have made an error at 5:57 where you _should_ have obtained
(4b² + 4b + 1)(b² + b + 1) = 1
which has the real solutions b = 0 and b = −1.

[@ruud9767]

z=1 is an obvious solution
Bring everything to the lhs: z^2 + (i-1)z -1=0
divide lhs (using long division) by z-1, since z=1 is a solution
you get z+i+1, so you can factor
z^2 + (i-1)z -1 = (z-1)(z+i+1) = 0
This gives z=1 or z=-1-i

[@0over0]

I too am a fan of synthetic division!

[@MichaelRothwell1]

Nice, I did it that way too! Pretty much like the 3rd method in the video except that there the sum of the roots =i provided a shortcut.

[@b.haroun]

A mistake@5:57 it equals to 1 not 0

[@bobbyheffley4955]

I used the second method.

[@mcwulf25]

Nice #3.

[@scottleung9587]

Got 'em both!

[@arekkrolak6320]

Nice solution. It transforms simple quadratic equation into cubic :)

[@aplusbi]

Yes, you are right 😜

[@SweetSorrow777]

You had a product of 2 quadratics which you have set to zero, which means you could've used the quadratic formula.
4th method? Writing z and the righthand side in polar form.

[@NadiehFan]

No, not really, because he made an error at 5:57 where he should have obtained
(4b² + 4b + 1)(b² + b + 1) = 1
which has the real solutions b = 0 and b = −1.

[@mcwulf25]

"I always make mistakes " 😂😂😂
I spotted the a^2 error and started writing thus message even I saw you correct it 😂

[@aplusbi]

😂😂😂

[@mtaur4113]

I had a solution by inspection and a little bootstrapping for a second solution. Method will work any time you have one of two roots in advance, or any time reducing the degree is helpful but you don't want to perform division.
z=1 by inspection.
Next, try z=(1+alpha), expand and cancel "P(1)=1+i" from the equation. Solve for alpha. A copy of alpha factors out and the rest is linear.

[@kmsbean]

at 6:24, can you not just factor 4b4+8b3+9b2+5b+1=0 into (4b2+4b+1)(b2+b+1)=0 which is 2 quadratics that are easily factorable?

[@E.h.a.b]

z^2 + i z = 1 + i
z^2 + i z - 1 - i = 0
z^2-1 + i(z-1) = 0
(z-1)(z+1)+ i(z-1) = 0
(z-1)(z+1+i)=0
z = 1 OR z = -1-i

[@aplusbi]

I should’ve known! Nice

[@mahoremujini]

Variante de la méthode 3:
z^2 -1=i-iz donc (z+1)(z-1)=-i(z-1)
D’où z=1 ou z+1=-i
S={1;-1-i}

[@0over0]

Definitely the shortest derivation we've seen yet.