This Just Can't Be Real

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Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.

#math #brithemathguy #i^i

[BriTheMathGuy]

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[mangodale.bingleman]

I love the fact that a imaginary number just turned into a real number.

[iamthebearerofchrist]

you literally brought our imaginations to life( turned an imaginary number to a real number), thanks a lot

[shaunthedcoaddict]

Wow, something fun to watch that I don’t understand but appreciate the math!

[ricobenning]

Kinda funny over the last weeks it seems like you’re uploading all the problems our complex analysis prof is handing to us haha
Great video!

[danmimis]

There's a shorter (one line) solution:
e^(ipi) = -1 now raise left and right to the power i/2 and you get i^i = e^(-pi/2)

[abhigyakumar]

We can do this simply putting value of i=e^(i*pi/2)
Then i^i=e^(i^2*pi/2)
So i^i=e^(-pi/2) {since i^2=-1}

[birajbbx]

The fact that he is doing maths but I cant see a single number the whole video scares me

[dariolazzari]

I think it's much easier this way:
i^i = (e^(iπ/2))^i = e^(π/2 * i^2) = e^(-π/2)

[angelmendez-rivera]

Technically, we regard the complex logarithm as ill-defined. It just makes no sense as a concept that extends the real logarithm, for many issues. Coterminal angles are not the only issue. Contour integration makes this much more damning. That being said, the function Log(z) = ln(|z|) + atan2[Im(z), Re(z)]·i can be a useful one to define in some limited contexts, and since it is mildly analogous to the logarithm, the notation has stuck on, despite it being somewhat misleading. However, I think it would be healthier to avoid thinking of the above function as an actual logarithm. Superficially, it has some of the same properties, but it causes more conceptual problems than it solves.

Anyway, if you interpret the notation ζ^ψ specifically as referring to exp[ψ·Log(ζ)], then it makes perfect sense to think of i^i as being equal to exp(–π/2). However, this interpretation of the notation is inconsistent with how exponents are typically treated. For example, it makes powers with base 0 undefined, and it also changes the odd fractional powers. Keeping in mind this inconsistency is something that many people need to be reminded of.

[TheStumbleGuysPlayer]

1:49 that r is still stuck inside of that ln

[omerbar]

The title
Was so accurate
I yelled wtf when I realized what was going on and my mom came to see if everything's fine lol
amazing video

[emilyscloset]

I didn't even bother taking the natural log.

I knew z=re^i*theta
hence, i = e^(i*pi/2)
thus i^i = (e^(i*pi/2))^i
=> i^i = e^(i*i*pi/2)
=> i^i = e^(-pi/2)

[samw]

I'm glad we see eye to eye on this one.

[Asterisme]

z-->i^z is a multivalued function. You can't attribute a unique value to i^i.

[OptimusPhillip]

I personally just substitute the i in the base for e^(i*pi/2), raise that to the ith power, replace i*i with -1, and get e^(-pi/2)

[shubhsrivastava]

A real number (-1) raised to the power of a real number (0.5) gives an imaginary number (i) and raising an imaginary number (i) to the power of an imaginary number (i) gives a real number. Wow!

[MonzennCarloMallari]

"i to the i, it's about a fifth" - Matt Parker of the legendary Parker Square

[Luc]

Wouldn't it be easier (and maybe less insightful) to substitute i by its polar form, exp(-i*pi/2)?
Then you'll quickly find i^i = exp(i*-i*pi/2)=exp(pi/2)

Edit: I missed a minus sign! Thanks for noticing!

[shreyhaansarkar]

I don't exactly understand the step he takes at 2:34. How does ln(i) = i(pi)/2 turn to e^i *i(pi)/2?