Square root i
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- เผยแพร่เมื่อ 15 ก.ย. 2024
- Have you ever wondered what the square root of i is? Is it real or complex? Watch this video to find out!
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Here I am taking the principal square root, where the angle is between 0 and pi. There is another root, which is -sqrt(2)/2 -isqrt(2)/2
I guess
-√2/2 + i√2/2 and
+√2/2 - i√2/2 are roots too
Since total there are supposed to be four 4th roots of negative unity
What’s is a non-principal root in general?
there are infinetly many roots for this solution because you can go n-times around the circle and still land on the same point with the same answer.
Swap e^((i•π):2) with
e^((i•π•2πn):2) n element of Z
@@rabindranathghosh31 Those are the values for √-i
@@rabindranathghosh31 No. There are 4 fourth roots of -1, but not all four of them are 2nd roots of i. Two of them are 2nd roots of i, and two of them are 2nd roots of -i.
I feel the need to point out that this claim can be immediately vetted as false, because it contradicts the Fundamental Theorem of Algebra; your claim is that the 2nd degree polynomial x^2 - i has four roots.
A scenic route alternative, without using Euler's equation:
(a + bi)^2 = i
a^2 + 2abi - b^2 = i
Hence a^2 - b^2 = 0, and 2ab = 1, so a = b.
So 2a^2 = 1, so a = +/- sqrt(2)/2, so b = +/- sqrt(2)/2.
So, the roots are sqrt(2)/2 + sqrt(2)/2i (the principal square root) and - sqrt(2)/2 - sqrt(2)/2i.
Omg hi Epic Math Time!!!
a^2 - b^2 = 0 only means that |a| = |b|, not a = b.
@@drpeyam Hello! How's it going Dr. Peyam?
@@chemistro9440 Yes, a^2 - b^2 = 0 tells us that a and b differ by at most sign, but 2ab = 1 tells us that they must be the same sign.
You're just brilliant 😀😀. Regards from India.
Makes sense. This is a good one to think about in terms of how multiplication of complex numbers gives a product of magnitudes and sum of angles.
You can also do it this way
i=i
2i=2i
2i= 1-1+2i
2i=1+i^2 + 2i
2i=(i+1)^2
(2i)^1/2= i + 1
i ^1/2=(i+1)/2^1/2
You are legend . Please make a video on Reiman Zeta function and Mock theta function.
i th root of i: th-cam.com/video/YGyLpm44Dpo/w-d-xo.html
ln(i): th-cam.com/video/asPo4xsxFlM/w-d-xo.html
square root ln(i): th-cam.com/video/fckRSvrbfws/w-d-xo.html
Your video makes me a genius in front of my class mates and teacher🥳
You could have multiplied it by 2/2 then add i^2 and 1 to make a perfect square
If you know the complex plane then it is immediate to see that sqrt( i ) = exp( i 90º)^½ = exp ( i 45º) = (1 + i)/sqrt(2) or the opposite.
One cool short proof you could show is for cos(a+b)=cos(a)cos(b) - sin(a)sin(b) taking a detour through the complex plane exploiting that cos(a+b)=Re{e^i(a+b)}.
I love that, thank you!
Nice to see you're trying #Shorts too! I'll be posting one later today.
Easy method without e or pi or sin or cos. Multiply inside square root:
ix2/2. 2i is a perfect square, square root of 2i is i+1. This over square root of 2, done.
How did you get sqrt(2i) = 1+i, looks like circular reasoning
@@drpeyam ?? You never took (1+i) to various powers? This is just what mathematicians do, kind of like knowing the powers of 2. Anyway, what is (1+i)^2? Yep, you get 2i.
Oh ok that makes sense, but it assumes that you know which one to square
@@drpeyam That's what I mean. You're more into analysis. Maybe it's more of a number theorist and algebraist thing. You know, sort of like the Euler sum of the reciprocal of squares. You see that in a problem and automatically know the sum, and continue from there. Us number guys automatically know that when we see 2i, it's a petrect square. Same with other numbers, like (sqrt(2) +1)^n, sqrt(n), etc. We immediately recognize certain properties of the quantities. When I was a kid, I taught my little brother some math tricks. We used to simultaneously recite powers of 2 into the hundreds of millions just for fun, LOLOL.
(1+i)^2 = 2i
So, i = (1+i)^2 / 2
So, sqrt(i) = +- (1+i)/sqrt2.
Is there any loop hole?
Nope
I love your videos ur a fantastic teacher I hope you get 1 million subs because you deserve them
I'm always open to subscribe and support educational chanmels
And I'm always looking to get help from online education channels
Actually, there's another way. But the answer you got is also correct.
Yo había llegado hasta el tercer paso, de e^(iπ/4) y me había quedado estancado porque de trigonometría sé poco, aunque me olía a que había que usar trigonometría.
Geometric interpretation of the problem would take 10 seconds at most. Draw a circle, draw the point at pi/2, then another point at pi/4, and done!
That rationalisation in end indicates the teacher in you
I remember there were actually two valid roots!
i like these short vids
Did you rely on a commutative law for merging real powers which isn't available using complex numbers? And converted sqrt to a power of half?
I didn’t use any commutative law
I have got an another way
U can let sqrti = a+bi and thens solve for a and b. But tbh, this was way better because it was fast 🤩👌💕
This is the glory crown of Math beauty !
That was a great path
Can you elaborate about following - if we do sqrt(4), there is only one root, e.g. 2 (and -2 is not root). Why complex sqrt have multiple roots?
Both 2 and -2 are second roots of 4. When you write sqrt(4), you are specifically referring to the positive one, the principal square root. That is, sqrt(4) = -2 is false, but not because -2 is not a second root of 4; it's false because of the notational definition of sqrt(4).
The idea of a principal root is a solid concept for real roots because "positive" actually has a concrete meaning: it's greater than 0. The idea of a principal square root for complex roots is shaky at best, because C does not have a total order; an arbitrary complex number cannot be said to be "greater than 0."
Can this be done without using the Euler notation, and just using polar form? I think we can, but it takes a little bit more time.
this can be done intuitevely. if sqrt(-1) is i and that corresponds to a 90 degree rotation around the complex unit circle then sqrt(i) corresponds to a 45 degree rotation
Black pen red pen did this in 9 monutes or something around that
😂
Minutes*
√ī speedrun 00:00:31
How about, instead of taking the 2nd root of i, take the i-th root of 2?
edit: I think the algebra shown in the video will help us find the answer to this question too, but what does that _i-th root of 2_ really mean?
I think that'll be cos(ln2) - isin(ln2)
@@rabindranathghosh31 So you think (e^(-iln2))^i=2 ?
.. well let's calculate: -i*i=1, 1*ln2=ln2 and e^ln2=2 so I agree.
If we want the ith root of 2, we have 2^(1/i) = 2^(-i) = e^ln(2^(-i)) = e^(-iln(2)) = cos(-ln(2))+isin(-ln(2)).
As for what raising to the ith power means, this is something I have recently thought about briefly. It appears in some cases that as multiplication by i is a rotation by 90 degrees (tracing a path along a circle), exponentiation by i has a geometric relationship to a cardioid.
-1 = e^iπ am i right then y u wrote π/2
I understood sir....
Square root of i is -1
No?
@@drpeyam why sir
@@drpeyam the imaginary part of Sqrt of i is -1
that was a weird background music
ooh now do cos(2 pi / 5)
Wow that's a new method
@@mathsandsciencechannel exactly
@@mathsandsciencechannel this one is amazing
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Seems legit
Papa flammy would be proud
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