No video
Solving A Locus Problem | Problem 279
Рет қаралды 1,393
Ай бұрын▶ 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.
❤️ ❤️ ❤️ 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!!!
❤️ This is Problem 279 on this channel!!! ❤️
🤩 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 @KZfaq @Apple @Desmos @GoogleDocs @canva @NotabilityApp @geogebra
@chaosredefined3834 Ай бұрын
|z / (z + i)| = 1 |z| / |z + i| = 1 |z| = |z + i| |z|^2 = |z + i|^2 Let z = a + bi |a + bi|^2 = |a + (b+1)i|^2 a^2 + b^2 = a^2 + (b+1)^2 a^2 + b^2 = a^2 + b^2 + 2b + 1 0 = 2b + 1 b = -1/2 So, z = a - i/2 Verification: |(a - i/2) / (a - i/2 + i)| |(a - i/2)/(a + i/2)| |(2a - i)/(2a + i)| |2a - i| / |2a + i| sqrt(4a^2 + (-1)^2) / sqrt(4a^2 + 1^2) sqrt(4a^2 + 1) / sqrt(4a^2 + 1) Which is 1, as 4a^2 + 1 can never be zero. So, this is true for all values of a.
@ProficiencyMusic Ай бұрын
I did it in my head and got -i/2
@kpt123456 Ай бұрын
Congratulations
@tolkienfan1972 Ай бұрын
But there isn't only a single answer.
@mcwulf25 Ай бұрын
@tolkienfan1972 you sure?
@tolkienfan1972 Ай бұрын
@@mcwulf25 did you watch the video?
@lackadaisicalll Ай бұрын
a-i/2
@sail2byzantium Ай бұрын
Well. I am glad somebody understood this . . . I lost the thread around 6:18 and why you square-rooted everything. And for such messy math, you should slow it down a bit. Things felt a bit rushed as to what exactly you were doing and why you were doing it.
@aplusbi Ай бұрын
Thanks for the feedback
@scottleung9587 Ай бұрын
I also got z=-i/2, corresponding to the line y=-1/2.
@JayTemple Ай бұрын
I got all numbers where the imaginary part is -1/2, but I didn't feel confident that there weren't other solutions.
@tolkienfan1972 Ай бұрын
z=-i/2 is a point, not a line
@phill3986 Ай бұрын
😊😊😊👍👍👍
@sdspivey Ай бұрын
3rd method: Cross multiply, ±Z = Z+i Subtract Z from each side, so 0=i or -2Z=i First is not possible, so Z=-i/2 Just a single valid answer, not a range.
@user-fz9go8pj4t Ай бұрын
This is totally wrong!!! You are taking the absolute value like a real number and you coincidentially get a part of a correct answer, but this is unacceptable
@sdspivey Ай бұрын
@@user-fz9go8pj4t Then show me another value that works.
@Nobodyman181 Ай бұрын
xy-xy=0
@mcwulf25 Ай бұрын
Yes, why didn't he cancel these and make that imaginary coefficient simpler? 9:51