Can You Solve A Nice Exponential Equation? | Problem 235

▶ 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/shop/sybermath
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 235 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.org/wiki/Complex...
/ @sybermath
/ @shortsofsyber
#complexnumbers #aplusbi #jeeadvanced #jee #complexanalysis #complex #jeemains
via @KZbin @Apple @Desmos @GoogleDocs @canva @NotabilityApp @geogebra

[@innovbrain]

I love your arithmetic solutions. This question was too easy. Only add Ln to both sides then you get 0 for ln(1), which is Z.lne=0, lne equals 1 already so it's Z that equals zero because 0*1=0

[@stevestarcke]

This channel is so wonderful! It always makes my day to see what looks insoluble is actually easy. Here, I can visualize myself as Ramanujan: A brilliant mathematician with little formal education.

[@aplusbi]

Thank you! 😍

[@prateek1.9]

somebody in comments really said, pls solve : (a+bi)²=b+ai

[@Nobodyman181]

Pls solve (a+bi)²=b+ai in video pls🙏

[@Nobodyman181]

Very easy

[@aplusbi]

Thanks a lot 😊

[@seanfraser3125]

Here’s the way that I like to do it that doesn’t depend on log (since that would be kind of circular reasoning). Let z=x+iy where x and y are real. Then 1 = e^(x+iy) = e^x*e^(iy) = e^x*cosy + ie^x*siny Setting real and imaginary parts equal, we have the following equations: e^x*cosy = 1 e^x*siny = 0 Since e^x can never be 0, the second equation gives siny = 0, or y = n*pi, where n is an integer. Plugging this into the first equation we have e^x*(-1)^n = 1 Since e^x > 0 for all x, (-1)^n = 1, so n is an even integer. We thus have e^x = 1, so x = 0 (x must be real). Putting this together, writing n=2k since n is even, we have z = 2kpi*i, where k is any integer.

[@trojanleo123]

I believe you are describing the "1st method" that was used in this video.

[@seanfraser3125]

@@trojanleo123 I usually give my solutions before watching the video so that I’m unbiased. They usually overlap with a+bi’s.

[@hafez591]

z=2.pi.n.exp(i.(pi/2+2.pi.k))

[@whyyat3470]

Umm... z = 0

[@aplusbi]

that's it?

[@leoheriban5809]

If an exp. equals 1 the exponent has to be 0

[@aplusbi]

that's it?

[@trojanleo123]

z = 2πni where n ∈ ℤ