The second one is easy:- e^(i*pi) = -1 ith root of e = e^(1/i) So (ith root of e)^(pi*i^2) = -1 So (ith root of e)^-pi = -1 Flip over fraction So (ith root of e)^pi = -1 We know e^(i*pi) = -1 So (ith root of e) = e^i That's a distance of 1 radian around the unit circle So (ith root of e) = cos(1)+i*sin(1) Or about 0.54030 + 0.84147i
@@dnpendown3199 Or use Qalculate, the best desktop and command line calculator available! > root(i,e), root(e,i) [root(i, e), root(e, i)] = approx. [0.8376315 + 0.54623573i, 0.54030231 - 0.84147098i]
@Smash Boy Yeah, technically speaking Complex numbers are just an algebraic estructure made from real numbers (just like matrices for example) so it makes sense that if you apply a function to an "imaginary" number (a number on the Y axis of the complex plane) it gives you a "real" number (a number on the X axis of the complex plane).
The reason why i^i = e^(-pi/2) is because i^i = e^(log(i)i) = e^((pi/2)i * i) = e^(-pi / 2). There are infinite answers because the log(z) has infinite answers; as it is defined as ln |z| + arg(z)i, where arg(z) is the angle (which can be, in this case, pi/2, 5pi/2, 7pi/2, ...).
This guy rekindles our faith in the fact that things that seem to be very tough, are actually very easy, we just need to take one right step, and everything else becomes so easy!
I did some other method : Consider i^(1/i) and get i = (-1)^(1/2) and -1=e^iπ so finally : i=e^(iπ/2) so in the end we have i^(π/2) as the "i"s cancel out !
You could actually just start with e^(ipi)=-1. Take the square root of both sides: e^(ipi/2)=i. Then take the ith root of both sides. e^(pi/2)= i^(1/i).
Another way to do this is using De Moivre’s theorem, where i on the complex plane would be pi/2, so theta is pi/2/i, so e^i theta would simplify to e^pi/2
Ognjen Kovačević u know. It's weird that even tho I teach the same way in my classes. But I have been doing different things on YT nowadays and I have been enjoying it way more also.
As mentioned before, I love this channel. Found it randomly, but so glad I did. Currently a grad student taking complex analysis (as an elective, took it 2years ago). All of your complex videos are just supplementary/review for me, but they're still very satisfying to watch. Got any plans for singularities?
The whole point of the question was to explain that e term, and you bring that from within the thin air, solving the sums fast doesnt mean you miss the real part of the solution
You can check its true, if you raise both sides to the power i, you get i = e^(i*pi/2) which is the complex polar form of a vector with argument pi/2 and magnitude 1, which is of course just i on the complex plane. The family of solutions are associated with all arguments that equal pi/2 mod 2pi, so any integer number of times around the units circle landing back at i.
That moment when you try to checkmate a grandmaster and he just literally flips the board. Great stuff man, I didn't this was a question that bothered me since 3 minutes ago but thanks for the answer anyways.
Sir there can also be my method: Let i^1/i=x Take ln both sides.. Now 1/i×ln(i)=ln(x)-----(1) Now as we know e^iπ=-1 And(-1)=i^2 e^iπ=i^2 then take ln both sides then (iπ)=2ln(i) Which equals (iπ)/2=ln(i)------(2) Now put (2) in (1)= {ln (i)=iπ/2} 1/i×iπ/2=ln(x)==== π/2=ln(x) so e^π/2=x I hope this can also be a solution...☺
These just make me so cheerful..... I have to limit myself to one a day ( and then I'll start again from the beginning because I'm still forgetting stuff like when I was a kid. )
Why so complicated though? Q: i^(1/i)=? Solution Rewriting the BASE i on the polar form, by analyzing the complex plane we get that r=1 and θ=π/2+2n, n∈ℤ. Therefore, i=cos(π/2+2n)+isin(π/2+2n). By Euler's formula this can be written as i=e^(i(π/2+2n)). Keeping the EXPONENT on it's original form, we can conclude that i=(e^(i(π/2+2n)))^(1/i)=e^(π/2+2n) by simplification A: i^(1/i)=e^(π/2+2n) By plugging in n=0 we get the answer i^(1/i)=e^(π/2), as presented in this video.
So much love given to - i = e^(iπ2(k+¼)) - in all of its forms and so little given to - i = -e^(iπ2(k-¼)) i^(1/i) = -e^(-π/2 + 2kπ) or e^(π/2 + 2kπ) You can't just add multiples of 2π to the positive result and declare victory.
damn I didn't know about imaginary numbers more than how to write them and how to show them on complex plain. yet after watching several other videos of yours I managed to get this one by myself.
is this really the right answer? look at this: i = exp(pi * i / 2) = exp(5* pi * i /2) but now the i-th root of it can be exp(pi / 2) or exp(5* pi /2) or any other power of the form (4n+1) * pi / 2
When I saw the video I took a few minutes to think about it for myself and then come back to see if I was right and I was. Thanks I like keeping my mind sharp with little math riddles like this one.
You could easily motivate this by analogy with a square root. Like square root of 4 can be found by finding a number a such that: a^2=4 ==>a=2. Similarly, the ith root of i means find an a such that: a^i=i ==> a=e^(pi/2)
I just realized that to get from i^i to i^1/i you need to multiply the exponent by 1/i^2 and that is 1/-1 which is the same as - 1 so (i^i)^-1 = i^1/i then you can solve it like this: (i^i)^-1 = (e^(-pi/2))^-1 = e^(-1*-pi/2) = e^(pi/2)
Let's see, this is math, where we always try to answer a new problem by casting it into terms of one we've already solved . . . i-root of i = i^(1/i) = i^(-i) = 1/(i^i) and we've already visited i^i; it's = e^(-½π). So the (principal) i-root of i = e^(½π) ≈ 4.810477...
Ian but that would require you know that e^(pi i/2) is equal to -minus one- i, and how do you know that? tho I guess here you have to know what i^i is so not really that much of a difference
ApplepieFTW actually e^(pi i/2) = i, but I catch your point. The problem is that roots and noninteger powers are defined in terms of the exponential function on C, so...
While verifying the solution to this (and to i^i), I encountered a paradox: Since i^i and i^(1/i) are reciprocals of each other, they should multiply to be equal to one. i^i*i^(1/i)=e^(-pi/2)*e^(pi/2)=1 However, if I evaluate i^i*i^(1/i) a different way, I get a different answer. i^i*i^(1/i) = i^(i+1/i)=i^((i+1)/i)=i^((-1+i)/-1)=i^(1-i) Using exponent rules, we can rewrite i^(1-i) as i*i^(-i)=i/(i^i) We have already calculated that i^i=e^(-pi/2), so we can substitute this into the original equation i/(i^i)=1 i/e^(-pi/2)=1 i*e^(pi/2)=1 The math used to evaluate i^i in the first place is clearly right, so I don't know what the problem is. How can a real number somehow be equal to an imaginary number?
I just checked wolfram alpha btw, and it says 1, so I'm assuming that math isn't broken. However, I am also strongly convinced that the alternative approach I took to evaluating i^i*i^(1/i) is also correct.
So this right here actually makes A LOT OF SENSE , if we build a graph of nth root of n , it peaks at e-th root of e , and i-th root of i is 4.81 that should be possible but this is imaginary so that's why we get a real numbers like that
Could one also write i^(1/i) as (e^(pi/2)i)^(1/i), because e^(pi/2) = i according to Euler's formula, allow the i and 1/i to cancel out, and take the remaining e^(pi/2) as a final answer?
I accidentally created a generalized formula for the i-th root of any complex number with a radius of 1 in school which consists of i-th root(z) = e^a, a is equal to the angle of the complex number in polar form. The first version of my formula is the i-th root(z) = i-th root(r) • i-th root(e^(i•a)), a is still the angle of the number in polar form (only works with the radius of 1).
Also, I removed the i-th root of r from the first version simply because it is useless since it is the i-th root of a number, where the i-th root is not defined. So r must be equal to 1 and the i-th root of 1 must be equal to 1.
Doesn't this give a problem? i=e^(pi/2+2 k pi)i for any integer k, thus we obtain e^(pi/2)*e^(2 k pi) for any integer k, thus we obtain a lot of different values?
Considering that on a Cartesian plane moving by pi is 180° so since it is pi/2 which is 90° therefore you can have the statement pi*(4n-3)/2 where n is any real number ;)
Why did you multiply by i over i? You could have substituted e to the i pi/2 as first step. And then cancel out the i and get the result in 2 steps. :)
Do you enjoy complex numbers and up for a challenge? If so, check out x^x=i th-cam.com/video/vCdChDmMYL0/w-d-xo.html
How is this practically applied to real life????
You can't solve subfactorial of i.
The most surprising thing about this is
I understood more than half of what he explained
I agree
I understood it all!
thats because hes a good teacher
Yeah and I'm terrible at maths.
0:32
"I don't like to be on the bottom, I like to be on the top"
Me: *scrolls down to comments to see if anyone made a joke about that*
A joke would be like what?
Super Psych that is literally exactly what I just did 😂😂😂 you made my day
I assume the top would be the video section, while the bottom is the comment section.
In the video it's already meant to be a joke.
Same here...
0:32 "That's what she said!" xD
*Learns about imaginary numbers*
TH-cam:
What's 1/i?
What's e^i?
What's i^i?
What's sqrt(i)?
What's log(i)?
What's i-th root of i?
What's sin i?
What's sin-1(2)?
What's i factorial?
What's integral of x^i?
刘颢云 lolll
Ya got lucky...
I haven't even learned about logarithms and have no clue what e is...
@@Ramu-10 you gotta love when you figure out whats going on
@@Ramu-10 it is euler's number, which approxamatly is 2.71.......
@@Ramu-10 then watch this amazing video about "e" and it's derivation.
th-cam.com/video/m2MIpDrF7Es/w-d-xo.html
"i" understand this a whole lot better now.
😂😂😂
lmao
😂
@Shri Hari's Animations it's not comedy it's komedi
“i” get it, even though the “roots” of the ha”i”r on my head are gray
Do the e-th root of i and the i-th root of e
The second one is easy:-
e^(i*pi) = -1
ith root of e = e^(1/i)
So (ith root of e)^(pi*i^2) = -1
So (ith root of e)^-pi = -1
Flip over fraction
So (ith root of e)^pi = -1
We know e^(i*pi) = -1
So (ith root of e) = e^i
That's a distance of 1 radian around the unit circle
So (ith root of e) = cos(1)+i*sin(1)
Or about 0.54030 + 0.84147i
just go on wolfram|alpha
Lyri Metacurl
Lol
@@anselmschueler wtf
@@dnpendown3199 Or use Qalculate, the best desktop and command line calculator available!
> root(i,e), root(e,i)
[root(i, e), root(e, i)] = approx. [0.8376315 + 0.54623573i, 0.54030231 - 0.84147098i]
“I dont like to be on the top. I like to be on the bottom” - blackpenredpen
lmao
Kinky
Wow, complex roots of imaginary numbers are real......math is so weird!
Sometimes they are, but not usually.
But bro real root of real numbers is also imaginary. For example square root of -1 😂
@Smash Boy Yeah, technically speaking Complex numbers are just an algebraic estructure made from real numbers (just like matrices for example) so it makes sense that if you apply a function to an "imaginary" number (a number on the Y axis of the complex plane) it gives you a "real" number (a number on the X axis of the complex plane).
because “One of the miseries of life is that everybody names things a little bit wrong.” - Richard Feynman 1985.
Purely imaginary in this case, imaginary root of imaginary...
The reason why i^i = e^(-pi/2) is because i^i = e^(log(i)i) = e^((pi/2)i * i) = e^(-pi / 2). There are infinite answers because the log(z) has infinite answers; as it is defined as ln |z| + arg(z)i, where arg(z) is the angle (which can be, in this case, pi/2, 5pi/2, 7pi/2, ...).
but why log(i)=(pi/2)i ?
J CO Because log(-1) = iπ and -1 = i^2
Or you could just
i = e^(iπ/2)
ith root both sides
i√i = e^(π/2)
+2kπ ?
@@arthurmoiret6076 That's the point when he said there are infinitely many solutions to this
I did that too!!!!!
@@arthurmoiret6076 or -2kπ
This guy rekindles our faith in the fact that things that seem to be very tough, are actually very easy, we just need to take one right step, and everything else becomes so easy!
I just stumbled upon your videos while procrastinating. You're a really warm and enthusiastic talker, keep it up sir! :)
I did some other method : Consider i^(1/i) and get i = (-1)^(1/2) and -1=e^iπ so finally : i=e^(iπ/2) so in the end we have i^(π/2) as the "i"s cancel out !
i blinked and he got pi out of seemingly nowhere
You could actually just start with e^(ipi)=-1. Take the square root of both sides: e^(ipi/2)=i. Then take the ith root of both sides. e^(pi/2)= i^(1/i).
Euler’s number and pi, name a more iconic duo
Another way to do this is using De Moivre’s theorem, where i on the complex plane would be pi/2, so theta is pi/2/i, so e^i theta would simplify to e^pi/2
I'm in love with mathematics now
I was in love with math. Then they said “do specialists”
I'm even more amazed that you can formulate these math questions with answers.
I just wasted 2 minutes being confused by a problem I never even knew existed and still don't understand
Damn, same, I clicked this in my recommended (I've always been horrible at math) I don't get how people in the comments get it so easily
Drink game:
Take a shot after he says the letter "i"
You'd get drunk after the first 10 seconds.
alternate method:
i^(1/i)=x
i=x^i
now we know e^(i . pi/2) =i
therefore i=e^(i . pi/2) = x^i
so, x=e^(pi/2)
Who thinks new BPRP is more interesting and better? *raises hand*
Ognjen Kovačević thank you!!!
yay!!
Ognjen Kovačević u know. It's weird that even tho I teach the same way in my classes. But I have been doing different things on YT nowadays and I have been enjoying it way more also.
blackpenredpen wait you're a teacher?! I thought you're just a super smart student!!
As mentioned before, I love this channel. Found it randomly, but so glad I did.
Currently a grad student taking complex analysis (as an elective, took it 2years ago). All of your complex videos are just supplementary/review for me, but they're still very satisfying to watch.
Got any plans for singularities?
Travis Hayes great idea! I will discuss this with peyam bc we are filming together this Friday
in a sense,all of his videos are complex
in two,actually. at least two
I don't believe a word of what you say! You doubtless are as flummoxed as the rest of us mortals. Quit showing off!
You are a math god
And you are invisible
Invisible, you scared me
the heck is happenin here????
Hangul Choseong Filler
I FINALLY FOUND YOU (only intellectuals understand)
This video is amazing. Thank you for the break down!
You're explanations are simple and crisp, really helping me learn mathematics better
Your*
i=e^i pi
i'th root of i =( e^i pi/2) ^1/i= e^pi/2
There are other solutions as well: (e^(pi/2+2*n*pi))^i = i for integers n.
black mic red shirt yay!
Imagine if imaginary roots of other imaginary things could be real. Like trees.
e, pi, and i: The Zeus Poseidon and Hades of math
Do Aid correct name of gades?
I don’t even know how to begin to do this
I'm going to use this information when I file my taxes.
That was the question that i had not been able to solve in my exam thanks
same😢
I still find it odd that "i" with absolute magnitude of "1" can ever twist itself into something outside of that, as in > 1.
Based
I love it when blackpenredpen features blackpenredpen.
You are the most brilliant math person I have ever come across. I am in awe.
The whole point of the question was to explain that e term, and you bring that from within the thin air, solving the sums fast doesnt mean you miss the real part of the solution
Tell them we wrote "i" as "e^i(theta)" and the "i" and "1/i" cancels out
That ball that you always keep in your hand makes you look like an ood from DW xD
You can check its true, if you raise both sides to the power i, you get i = e^(i*pi/2) which is the complex polar form of a vector with argument pi/2 and magnitude 1, which is of course just i on the complex plane. The family of solutions are associated with all arguments that equal pi/2 mod 2pi, so any integer number of times around the units circle landing back at i.
That moment when you try to checkmate a grandmaster and he just literally flips the board. Great stuff man, I didn't this was a question that bothered me since 3 minutes ago but thanks for the answer anyways.
Another way is to get this result is to convert i into it’s polar form, you get there a lot quicker
(X^2-5x+5)^(x^2+7x+12)=1
Find all real values of x.
There will be 6 values of x.
Love from India.
nicely done.... the use of multi color to force focus reminds me of a math professor in the 1960's at the University of Maine.
Thank you Mike!
Your voice is a lullaby 😭❤️
And 5 years later: Just go straight for Eulers identity: i'th root of i = i^(1/i)=e^(πi/2i)=e^π/2 From 5 steps to 3 steps
Sir there can also be my method:
Let i^1/i=x
Take ln both sides..
Now 1/i×ln(i)=ln(x)-----(1)
Now as we know e^iπ=-1
And(-1)=i^2
e^iπ=i^2 then take ln both sides then (iπ)=2ln(i)
Which equals (iπ)/2=ln(i)------(2)
Now put (2) in (1)=
{ln (i)=iπ/2}
1/i×iπ/2=ln(x)====
π/2=ln(x) so
e^π/2=x
I hope this can also be a solution...☺
These just make me so cheerful..... I have to limit myself to one a day ( and then I'll start again from the beginning because I'm still forgetting stuff like when I was a kid. )
Why so complicated though?
Q: i^(1/i)=?
Solution
Rewriting the BASE i on the polar form, by analyzing the complex plane we get that r=1 and θ=π/2+2n, n∈ℤ. Therefore, i=cos(π/2+2n)+isin(π/2+2n). By Euler's formula this can be written as i=e^(i(π/2+2n)).
Keeping the EXPONENT on it's original form, we can conclude that
i=(e^(i(π/2+2n)))^(1/i)=e^(π/2+2n) by simplification
A: i^(1/i)=e^(π/2+2n)
By plugging in n=0 we get the answer i^(1/i)=e^(π/2), as presented in this video.
So much love given to -
i = e^(iπ2(k+¼))
- in all of its forms and so little given to -
i = -e^(iπ2(k-¼))
i^(1/i) = -e^(-π/2 + 2kπ) or e^(π/2 + 2kπ)
You can't just add multiples of 2π to the positive result and declare victory.
I don't even think about this kind of stuff. It's crazy that some dude was like, oh there's an i'th root of I, let's solve what it is...
If I have learned anything from this channel it is that if you something strange to i, e and/or pi are going to pop out of the equation.
As far as I know, the transition from sqrt to power imposes restrictions on the argument. And here it is observed?
Well, _i_ root of _i_ is pretty easy to calculate.
The _a_ root of _b_ can be rewritten as _a_ ^(1/ _b_ )
So, _i_ root of _i_ = _i_ ^( 1/ _i_ )
Now, 1/ _i_ = -i. You can figure that out geometrically / visually with ease.
_a_ ^ ( - _b_ ) = 1 / ( _a_ ^ _b_ )
So, _i_ root of _i_ = _i_ ^( 1/ _i_ ) = _i_ ^ (- _i_ ) = 1/ ( _i_ ^ _i_ )
To calculate _i_ ^ _i_ , we can use one of euler's formulas. For starters, _a_ ^ _b_ = _e_ ^( ln( _a_ ) _b_ ), thus _i_ ^ _i_ = _e_ ^( ln( _i_ ) _i_ )
ln( _a_ + _bi_ ) = ln( length( _a_ + _bi_ ) ) + angle( _a_ + _bi_ ) _i_ , so _e_ ^( ln( _i_ ) _i_ )
= _e_ ^( ( ln( length( _i_ ) ) + angle( _i_ ) _i_ ) _i_ )
= _e_ ^( ( ln( 1 ) + angle( _i_ ) _i_ ) _i_ )
= _e_ ^( ( 0 + angle( _i_ ) _i_ ) _i_ )
= _e_ ^( ( angle( _i_ ) _i_ ) _i_ )
= _e_ ^( ( 1/2 _pi i_ ) _i_ )
= _e_ ^( 1/2 _pi i i_ )
= _e_ ^( -1/2 _pi_ ) becuase _i i_ = -1
Now, _e_ ^( -1/2 _pi_ )
All of that says _i_ ^ _i_ = _e_ ^( -1/2 _pi_ )
We still have to find 1/that.
1/( _a_ ) = _a_ ^(-1), so 1/( _e_ ^( -1/2 _pi_ ) ) = _e_ ^( -1/2 _pi_ ) ^(-1). Stacked exponents multiply, so _e_ ^( -1/2 _pi_ * -1 ). The negative cancel, resulting in:
_i_ root of _i_ = _e_ ^( 1/2 _pi_ ). The last part here can be calculated.
In conclusion, the _i_ root of _i_ = _e_ ^( 1/2 _pi_ ) ~~ 4.8105
You say "1/ i = -i. You can figure that out geometrically / visually with ease." But I fail to see that visually... I only get that algebraically
damn I didn't know about imaginary numbers more than how to write them and how to show them on complex plain. yet after watching several other videos of yours I managed to get this one by myself.
is this really the right answer?
look at this:
i = exp(pi * i / 2) = exp(5* pi * i /2)
but now the i-th root of it can be
exp(pi / 2) or exp(5* pi /2) or any other power of the form (4n+1) * pi / 2
When I saw the video I took a few minutes to think about it for myself and then come back to see if I was right and I was. Thanks I like keeping my mind sharp with little math riddles like this one.
but the real question is... what is i-th root of i based on Wolfram Alpha?
Max Haibara lol!!!!
Just tried it... it says what he said
I just did it before reading the comments, then I did e^(pi/2), then I compared the decimals, and they are the same, so W|A agrees.
@@MattMcIrvin The joke is that you don't need Wolfram Alpha to solve seemingly difficult equations.
Even calculators can compute it, and they give the same answer. (Done on a TI-84)
You should probably also update this version with e^(pi/2+2*pi*n)
"What we do now? We do our usual business" if math was that easyyy
By the fundamental theorem of algebra, there are exactly n roots of nth power. So, there are exactly i roots of i :)
Black pen blue pen would make it easier for my eyes to read. Thank you so much for your videos.
“I dont like to be on the bottom, i like to be on the top”
- pen guy
Of cooourse, (e^(pi÷2))^i is just equal to e^(i(pi÷2)) which is i in polar form.
The x-th root of x is at maximum at x=e for x a real number. e-th root of e is about 1.444... This tops it by 3.3...
I've given this as an interview question (I'm an EE) - it's easy if you just work your way through it.
This is so fantastic and beautiful. Thank you for sharing!!
You could easily motivate this by analogy with a square root. Like square root of 4 can be found by finding a number a such that:
a^2=4 ==>a=2.
Similarly, the ith root of i means find an a such that:
a^i=i ==> a=e^(pi/2)
The fact that it ends up a real number is really weird for some reason.
" I don't like to be on the bottom, i like to be on the top " (0:31)
- blackpenredpen 2017
0:30 "I don't like to be on the bottom. I like to be on the top."-Blackpenredpen, 2017
Wait ur a teacher!?
Philip Yao yes i am
blackpenredpen you're so young that we thought you're a 20-years old math youtuber haha
@@blackpenredpen where do you teach I have to go there (つ°ヮ°)つ
Nop, he is a math god.
@@victorvega8061 UC Berkeley
The above i^(1/i) has many infinite solutions e^(-pi/2) is one of it
Tak na dobry sen. U mnie luzik. Dobranoc, kurde jeszcze jasno.
such an interesting edge case!
There are infinite solutions: e^(pi/2 + 2*pi*n)
I just realized that to get from i^i to i^1/i you need to multiply the exponent by 1/i^2 and that is 1/-1 which is the same as - 1 so (i^i)^-1 = i^1/i then you can solve it like this: (i^i)^-1 = (e^(-pi/2))^-1 = e^(-1*-pi/2) = e^(pi/2)
i = e^i(2n+1)π/2 So i^(1/i) = e^(2n+1)π/2 for any n as an integer
Let's see, this is math, where we always try to answer a new problem by casting it into terms of one we've already solved . . .
i-root of i = i^(1/i) = i^(-i) = 1/(i^i) and we've already visited i^i; it's = e^(-½π).
So the (principal) i-root of i = e^(½π) ≈ 4.810477...
What would an ith root(x)graph look like?
_this just sounds like euler's formula with extra steps_
It used Euler’s formula at the end to compute it
i=e^(i*pi/2)
raise both sides to the power of i and you get
i^i=e^(-pi/2)
Could also just write e^(pi i/2) = i and take the root of both sides.
Ian But that requires you to know sqrt(i).
Obinna Nwakwue I meant the i-th root.
You have i = e^(pi i/2)
rt_i (i) = rt_i ((e^(pi/2)^i) = e^(pi/2)
Ian That's exactly what I thought of once I saw the last step.
This way is much simpler.
Ian but that would require you know that e^(pi i/2) is equal to -minus one- i, and how do you know that? tho I guess here you have to know what i^i is so not really that much of a difference
ApplepieFTW actually e^(pi i/2) = i, but I catch your point. The problem is that roots and noninteger powers are defined in terms of the exponential function on C, so...
Math is becoming a celebration for u people. Lots of love and respect from me.
While verifying the solution to this (and to i^i), I encountered a paradox:
Since i^i and i^(1/i) are reciprocals of each other, they should multiply to be equal to one.
i^i*i^(1/i)=e^(-pi/2)*e^(pi/2)=1
However, if I evaluate i^i*i^(1/i) a different way, I get a different answer.
i^i*i^(1/i) = i^(i+1/i)=i^((i+1)/i)=i^((-1+i)/-1)=i^(1-i)
Using exponent rules, we can rewrite i^(1-i) as i*i^(-i)=i/(i^i)
We have already calculated that i^i=e^(-pi/2), so we can substitute this into the original equation
i/(i^i)=1
i/e^(-pi/2)=1
i*e^(pi/2)=1
The math used to evaluate i^i in the first place is clearly right, so I don't know what the problem is. How can a real number somehow be equal to an imaginary number?
I just checked wolfram alpha btw, and it says 1, so I'm assuming that math isn't broken. However, I am also strongly convinced that the alternative approach I took to evaluating i^i*i^(1/i) is also correct.
"Let me do a REAL quick video for you" I see what you did there
So this right here actually makes A LOT OF SENSE , if we build a graph of nth root of n , it peaks at e-th root of e , and i-th root of i is 4.81 that should be possible but this is imaginary so that's why we get a real numbers like that
Could one also write i^(1/i) as (e^(pi/2)i)^(1/i), because e^(pi/2) = i according to Euler's formula, allow the i and 1/i to cancel out, and take the remaining e^(pi/2) as a final answer?
I accidentally created a generalized formula for the i-th root of any complex number with a radius of 1 in school which consists of i-th root(z) = e^a, a is equal to the angle of the complex number in polar form.
The first version of my formula is the i-th root(z) = i-th root(r) • i-th root(e^(i•a)), a is still the angle of the number in polar form (only works with the radius of 1).
Also, I removed the i-th root of r from the first version simply because it is useless since it is the i-th root of a number, where the i-th root is not defined. So r must be equal to 1 and the i-th root of 1 must be equal to 1.
Doesn't this give a problem? i=e^(pi/2+2 k pi)i for any integer k, thus we obtain e^(pi/2)*e^(2 k pi) for any integer k, thus we obtain a lot of different values?
A lot easear with the exponential formula, i^(1/i) = (e^i(pie/2))^(1/i)
I just thought of it as (e^(i*pi/2))^(1/i) then the i’s cancel out and boom e^(pi/2) this is only 1 step and a little easier
Using e^πi = -1, you can realize that i = √(-1) = √(e^πi) = e^(π/2)i
Now i^(1/i) is obviously equal to (e^(π/2)i/i), which is equal to e^(π/2).
0:30 the quote of all time
Considering that on a Cartesian plane moving by pi is 180° so since it is pi/2 which is 90° therefore you can have the statement pi*(4n-3)/2 where n is any real number ;)
Why did you multiply by i over i? You could have substituted e to the i pi/2 as first step. And then cancel out the i and get the result in 2 steps. :)
Omg, u did it so easily. The way that I've done was making:
Y=i^(1/i)
ln(Y)=ln(i)/i
then, i = e^(iπ/2)
ln(Y)=ln(e)×iπ/2i
ln(Y)=π/2
Y=e^(π/2)
00:32 Lost my concentration there
7
The very concept of an ith root doesn't make sense to me.
Natural log of (i) is equal to pi by 2 times i
tk i to lhs and tk it to power of the index and then raise it to the power of e