Finding The Absolute Value of z + 1 | Problem 255

▶ 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. ❤️ ❤️ ❤️
Recently updated to display "Books to Prepare for Math Olympiads" Check it out!!!

❤️ This is Problem 255 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

[bsmith]

Another approach to this problem. z=e^(i*t) can be seen as the graph of a circle centered at the origin with radius 1. Then z+1 is a translation along the x-axis.
The parametric form of z is (cos(t), sin(t)). Then z+1 has a parametric form of (1+cos(t), sin(t)). Then |z+1| is the distance from the origin to the point, which is sqrt[(1+cos(t))^2, sin(t)^2].
This gets us to where Sybermath is at 4:26 in the video. Then follow with the same trig manipulations to get |z+1|=2*|cos(t/2)|.

[iabervon]

Third method: let w be a square root of z with non-negative real part. If you recognize that |w| is 1 and 1/w is w-bar, you get: Since w has non-negative real part, that's 2cos(theta/2), assuming theta has been adjusted appropriately.

[wasimvillidad]

z + 1 = cisθ + 1 = cis(θ/2)[cis(θ/2) + cis(-θ/2)] = 2cos(θ/2)cis(θ/2) since cisα + cis(-α) = 2cosα
∴ |z + 1| = 2cos(θ/2) and Arg(z + 1) = θ/2. Too easy.

[SidneiMV]

|z| = 1

find |z + 1|

|z + 1| = 2 or |z + 1| = 0

(it's not the correct solution)