A tricky question that confused a lot of students | Complex numbers

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[@barbazzo]

Let's be clear here. 1^x=1, 1^x does not equal -4 under any conditions. The author of the video found the solution to x for e^(i*2*π*k*x)=-4, not 1^x=-4. Any solution for e^(i*2*π*k*x) = -4 would involve complex numbers, but the original equation 1^x = -4 is a contradiction for any real or complex x. The expression 1^x always equals 1 for any x, so there is no solution to 1^x = -4 in the real or complex number systems. Substituting 1 with e^(i*2*π*k) leads to complex solutions that do not satisfy the original equation, as 1^x remains 1 regardless of x.

[@jamesthechristian3951]

Hi, I really like all of your videos. You do a GREAT job of explaining your steps! I learn so much every time I watch your videos. I would love to see you work through Schrödinger's Equation if you ever want to tackle it?

[@hajstra1307]

Mister new Euler, just one example of one of your errors and its consequences. In in minute 10:00 you have: Ln(e^(i2kpi*x) = Ln(-4), (note the capital L for the complex log.) Applying the proper formula for the logarithm of complex number you will have: ln(1)+ i(2kpi*x + 2mpi) = Ln(-4) Do not apply ln(1) = 0. This is no longer true according to your algebra. The very 1st line of your presentation is 1^x = -4, after applying ln, and dividing by x, you have ln(1) = ln(-4)/x, and this is your algebra, erroneous, but you are bound to use this value of ln(1) all the time in your calculations. Also remember that in your algebra 1*1 is not equal 1, and 1*x is not equal x, and probably 2*x is not equal x + x. After substituting for ln(1) according to your algebra you have following to solve: ln(-4)/x+ i(2kpi*x + 2mpi) = Ln(-4) Also you can't remove the principal value of the complex number removing k = m = 0. There is no complex number w/o its principal value.

[@TUZZ5000]

"Mister new Euler" © Ahah😂 Golden!

[@peterjansen4826]

Complex numbers are not complex but calculations with complex numbers definitely tend to get complex. A lot of writing gets involved and you easily make a sloppy mistake.

[@jamesbond_007]

I think the k = 3 and 4 parts are unnecessary -- you don't really need to establish the pattern this far. If you must, use k in {1,2}, but that's all. Going to 3, 4, 1000 is unnecessary.

[@deinauge7894]

well... if someone asks, what is 1^(1/2 - i/pi ln(2)) would you really says it's -4? Because 1^x = 1 for all complex x, not only for real x. only of you add branches from 1^x=exp(x ln(1)) you CAN get a multivalued function - without any branch cuts!!. but usually 1^x = 1 and nothing else

[@TUZZ5000]

WolframAlpha believes that if you raise 1 to the power you got in your answer, you get exactly 1. Not -4 as would be expected for any value of x.

[@richardmullins44]

wow. You have certainly caught my attention. I noticed the other day from looking at these maths questions, that log(-2) has an infinity of complex values. "I don't remember learning that at school". Maybe I misremembered what I read the other day. but whatever it said, it has prepared me have a look at your post for this question. Now I'm stuck on 1^x = -4 take logs of both sides x log 1 = log (-4). I was prepared from what I read last week, to accept that log(-4) has exotic values. But how can x log 1 be anything else but 0?? I will watch your post anyway to see if it helps. I know I often make mistakes, but open ai agrees with me. Even open ai sometimes makes mistakes, and it often makes mistakes on difficult problems.

[@dieuwer5370]

No need for variable "m". Substitute left side with e^ (i2kπx), substitute right side with: 4e^(i(π+2kπ)).

[@AvihooI]

well the solution has two degrees of freedom, no? so you do need two variables

[@RealQinnMalloryu4]

(1)^2=1(x ➖ 1x+1 ). 2^2 (x ➖ 2x+2).

[@yusrialtamimi1570]

OMG ...... endless loops of complex numbers

[@Sauvenil]

Somehow I never learned about logarithms between high school and college. I figured that was part of this, but I don't know how to use them.

[@hajstra1307]

Do not warry, the author of presentation don't know how to use logarithms of complex numbers.

[@brian554xx]

At 5:43 I pointedly ask why anyone would not expect the thame anther for all k.

[@epd807]

Your videos are very informative! Thank you for making them. Also, what is your accent?

[@hajstra1307]

My advice: DO NOT REMAIN MISINFORMED by this video

[@epd807]

@@hajstra1307 I just was asking your nationality. You have a very unique accent.

[@hajstra1307]

@@epd807 I an not the author of this video

[@paulor.r.correia1789]

A inveja é uma merda

[@Dannyftv]

Hello i am the first.