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We know the principal square root of -1 is the imaginary unit i but what if we have the i-th root of -1? We will do a quick review on converting a complex number from the standard form a+bi to its polar form re^(i*theta). Then we will see the surprising answer that the i-th root of -1 will actually give us.
Related videos:
the proof of the Euler’s identity (with the Taylor series):
i-th root of i:
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0:00 what’s the i-th root of -1
5:30 check out Brilliant
6:08 challenging question (-1)^pi
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x^a = b (a,b,x complex numbers) in general has multiple solutions for x, sometimes infinitely many (e.g. when a=i like in this case). How is "the" answer for b^(1/a) selected?
challenge:i^sqrt(i)^sqrt(2)
红笔黑笔老师居然有英文频道
Why have you stopped making regular videos?
Hello bprp
The answer of minus one to power π
Equal e to power i
But if you do the same with the square root of -1: sqrt(-1) = (exp(i*pi*(1+2n))^0.5 = exp(i*pi*(0.5+n)), and it's either =i or =-i 🤔
Congrats on hitting 1 million subs!!
love your videos 🙂
So technically, I can say that sqrt(-1) is close to positive infinity? Neat. Calculating complex numbers in the real world can give such funny answers
Did this one in my head, lol. Just because of the famous e^(i*pi)=-1, so so the i-root of these gives is e^pi (or e^(pi+2pi*z)
way complex and imaginary problem
AAh!! after thinking about it, this is how you would represent a series or a sequence of real numbers.
Beautiful!! now.. what is the physics related to this number? or what does this mean in real world? thanks. I understand that you are a Math teacher, but as you already know math cannot be separated . that's the reason behind my question.
why r= 1?
bro wtf
Hey bprp, can you solve the integral of (e^x²√(x²-1)+1)/x²-√(x²+1)-1 ?
Hello ! I hope you see my comment
I saw this nice question so that I recommend it
The question is : solve the system of equations
a = exp (a) . cos (b)
b = exp (a) . sin (b)
It can be nicely solved by using Lambert W function after letting z = a + ib
Hope you the best … your loyal fan from Syria
Do i^pie
You lost me at 1:00
Very cool!
Cool!
Hello sir how are you? ,My name is Atiqur Rahman. I am a trader. I want to build up an AI Algorithm trading software system on trading, I have been trying to solve a math calculation on trading for a long time. I request your cooperation in this regard
A haven't seen the video, the answere is 1 right?
Couldn't you also say sqrt(-1) = e^(i(π/2 + 2nπ))
Does anyone know what i-th root of i is?
Isn't there technically only 1 solution to the problem, not infinite? I get that (-1)^(1/i) would have infinite solutions due to the argument being of the form pi + 2n*pi, with n being an integer.
However, since the initial statement used the radical symbol, it means the principal value only, right?
Respect bprp!
I used this method:
(-1)^1/i = -1^-i = -1
can anyone tell me whats wrong?
"we're all adults now, let's use pi". I'm dying xDDD
Thought-road: -1^1/i = i^i = 1
Bro the thing written on your shirt… Prove that
Fun. This time I did it before watching the video 😉
(-1)^π
We know e^(iz)=cos(z) + isin(z)
So
e^(iπ)=-1
Then we elevate it to π
(e^(iπ))^π=(-1)^π
So
(-1)^π=e^(iπ²)
But we know that with complex numbers e^(iz)=e^(i(z+2π))
Then
(-1)^π=e^(i(π²+k2π)), k=0,±1,±2…
In polar form: radius=1 Angle=π²+2kπ rad, k=0,±1,±2…
In binomial form
cos(π²)+isin(π²)
Real answer: i 8 sum π
If a root is a function, then it cannot have infinite results. What you gave is the result to the equation
z^i = – 1
Rather than
z=i th root of – 1
But still, for the rest, interesting!
Don't forget 2i×Peyam.
(-1)^pi = cos(pi^2) + i sin(pi^2) is the principal solution, but I think the general form would be (2n+1)pi^2 inside the cos and sin, n in integers, right?
BlackpenRedPenBluePen . . . YEAH!