For thsi problem it is a little bit faster to plug z=a+bi in at the beginning, beacause this way you can instantly show that a is 2 and that 2^bi is -1

@aplusbi

Ай бұрын

Very nice! I like your approach

The confusion I have with your presentation is sometimes you take only the principal solution and sometimes you takes all solutions. But I love this channel. For example in the root z + root(-z)= i ,you only took the principal solution(s). I'm not sure when one or the other is appropriate.

great video! which software do you use to make/draw them?

all you have to do is take log base 2 on both sides, expand the argument on the RHS, simplify log2(4), then we get to log2(-1) we know -1 = e^ipi, but pi is not the only angle that its valid, we can rotate 2pi so its e^ipi(1+2n) then its log2{e^ipi(1+2n)} you can do one of two: 1. you drop down the exponent 2. if you specifically want the solution in the video, use the change of base formula and ln MUST be used in this case because there is an e in the argument then you get the answer in the video in normal situations, we would not do this with a negative argument but since we are expecting complex solutions anyway its okay to do it in my opinion ahah i was right after all

Very cool 😎. Cheers from 🇨🇦.

@aplusbi

28 күн бұрын

Thanks 👍

Line of the day - With complex numbers, everything is possible!

@aplusbi

16 күн бұрын

😉

Where did the -4th power at 3:47 go?

Nice!

@aplusbi

Ай бұрын

Thanks!

z = 2 + i(2n+1)π/ln2

With complex numbers everything is possible: 1/0

@geraldsmith6225

Ай бұрын

Nooooooooo!

@aplusbi

Ай бұрын

😮😄

Please find x^(x^...)))..)=√i