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The neat thing is, the fact that there isn't a meta-imaginary means that our number system is closed.

@Kokurorokuko

2 жыл бұрын

Yeah. It's interesting that humanity had to spent so much time to make numbers "complete". There are quaternions, though, but they don't have some properties which imaginary numbers have.

@Noam_.Menashe

2 жыл бұрын

Does been able tomae undefined numbers and singularities still keep it closed?

@axoluna

2 жыл бұрын

It is still possible to define further numbers; even natural numbers are “made up.” Check out the dual numbers, a separate number system where the new unit E^2 = 0

@mertaliyigit3288

2 жыл бұрын

@@Kokurorokuko imaginary numbers also lack some properties of real numbers. Inequalities dont work anymore

@brandonklein1

2 жыл бұрын

Sets are only closed under particular operations, so we need to specify which operations and sets we're talking about for that statement to be meaningful.

Alternative method: Represent i in polar form Therefore, i = cos(π/2) + isin(π/2) By Euler's formula, e^(iθ) = cosθ + isinθ Therefore i = e^(iπ/2) Now √i = i^(1/2) = e^(iπ/4) Therefore √i = cos(π/4) + isin(π/4) √i = 1/√2 + i/√2 √i = (1+i)/√2

@PunmasterSTP

2 жыл бұрын

Yeah, that's what I thought about right when I saw the thumbnail.

@flowingafterglow629

2 жыл бұрын

@@PunmasterSTP Because it is so much easier

@gitgudnoobs7917

2 жыл бұрын

This is quite neat.

@angeluomo

2 жыл бұрын

Also my method. Had it in three steps.

@matejnovosad9152

2 жыл бұрын

So much uglier, but works

That’s pretty cool; personally, I like number 4.

@wockhardt9705

2 жыл бұрын

lmfaooo

@ktz1185

2 жыл бұрын

LMAO

@pasztesnik

Жыл бұрын

Perfection

@qwertyflags

Жыл бұрын

welp he's definitely not asian

@Bozitico

9 ай бұрын

SAME. Everyone always thinks it's weird, but it is important to solving math entirely. Well, every number has a role. But I like 4's role.😊😊😊

i was actually wondering this the other day. how this recommendation was so spot on blows my mind

@charliewallisch5581

2 жыл бұрын

Literally the same for me

@thedarkslayer9475

2 жыл бұрын

I was studying complex numbers and this got recommended

@wellen_good972

2 жыл бұрын

@@thedarkslayer9475 same

@benjaminkenney3706

2 жыл бұрын

Apple is listening through our phones I swear

@jacobw59

2 жыл бұрын

Same for me. I think the video was recommended based on our search histories and overall interests.

Technically we define i by having the property i^2=-1. That is, sqrt(-1) is the principal root of i^2. It's just more convenient for this video say i=sqrt(-1).

@createyourownfuture3840

2 жыл бұрын

blackpenredpen found the √i using both this method and using polar coordinates.

@themlaw9895

2 жыл бұрын

Nice video but i is not equal to sqrt(-1) : suppose sqrt(-1) exist, so sqrt(-1)² = sqrt(-1)×sqrt(-1) = sqrt(-1×-1) = sqrt(1) = 1; wich is different of i², which is -1 So i is not a number that has a value, his only property is i²= -1, so your starting point is wrong, nice try though

@kshitij7b286

2 жыл бұрын

@@themlaw9895 it is wrong √(-1)×√(-1)= √[(-1)×(-1)=√1=1 The property √a×√b=√ab holds if and only atleast one of the number is positive that is either a≥0 or b≥0

@createyourownfuture3840

2 жыл бұрын

@@themlaw9895 The property of exponents you just used isn't true for complex numbers, but you enter that field when you talk about √-1. i is as real as real numbers, and nature works in complex numbers, not real ones.

@MUJAHID96414

2 жыл бұрын

I am excited to learn upper class math🤩, but I haven't books

Just a small thing; 0:14 Imaginary numbers are a subset of Complex Numbers, not the same.

@user-sb9ho5jz3e

11 ай бұрын

before it was official, people used imaginary numbers to refer to any numbers that have an imaginary part

@juxx9628

9 ай бұрын

@@user-sb9ho5jz3e Official? Like what time? Cardano's time? Descartes time? Euler's time? Gauss time? Cauchy time? What is "official complex numbers"?

Another awesome video as always! I was curious about another method. Instead of using either Euler or De Moivre's formula, when you have the equation 0 + 1i = (a^2 - b^2) + 2ab*i, you could try setting a = b. If you do, then 1 = 2b^2 and so b = +/- sqrt(2)/2 = a, and sqrt(i) = +/- (sqrt(2)/2 + i*sqrt(2)/2). Knowing there are only two square roots of a complex number, you would be done. I also tried setting a = -b, and I got complex values for a and b, but plugging them into (a + bi) I got the same thing!

@BriTheMathGuy

2 жыл бұрын

Cool!

I was given this problem in an introductory maths module for physics and thought of a simple geometric interpretation. I figured √(i) as being half the rotation around the origin in an argand diagram from 1 to i, since multiplying by i has the effect of "rotating" a number 90 degrees counter clockwise. Therefore the coordinates for √(i) would be the points on the unit circle that were half way between 1 and i either way. This would gives two points on the unit circle, one going 45 degrees counter clockwise from 1, and another going 135 degrees clockwise from 1. You can quickly see these are both 45 degree angles from the axis, thus a^2 + a^2 = 1 --> 2a^2 = 1 --> a^2 = 1/2 --> a = +1/√(2) or -1/√(2) for the respective top right and bottom left quadrant solutions. Therefore √(i) = +1/√(2) * (1 + i) or -1/√(2) * (1 + i)

@sujaiy6646

2 жыл бұрын

Nice 😊👍👍👍

@kaan8964

2 жыл бұрын

What is the concept of rotating imaginary numbers called?

@andrewkarsten5268

2 жыл бұрын

@@kaan8964 it’s really more of the general interpretation of multiplication in general. It’s a great interpretation. I recommend 3blue1brown’s video on complex numbers, he gives a wonderful way to rethink what addition and multiplication actually mean. The way to think of complex numbers falls right out of that.

@Diaming787

2 жыл бұрын

@@kaan8964 complex numbers are in a form of a+bi but so is r*e^(i*theta), and theta is the rotation part. Example: i is equal to e^(i*pi/2) because it is rotated pi/2 radians from the +Real axis. -1 = e^(i*pi) because it is rotated pi radians, which is why when you multiply by negative numbers, you 'flip' to it's negative side.

you can also just know that a 45 degree, 90 degree, 45 degree triangle with hypotenuse of 1 has sides of 1/√2

@v44n7

Жыл бұрын

this, its a beautiful way to see it

@lathasri3483

2 ай бұрын

Amazing

Nice! I always thought of it as: i = e^(i pi/2) i^(1/2) = +- e^( i pi/4) and just leave it at that, because I prefer polar form for some reason

In the complex numbers, the nth root function takes on n values. Probably important to mention that.

@angelmendez-rivera351

2 жыл бұрын

No. Do you know what the definition of a function is?

@PubicGore

2 жыл бұрын

@@angelmendez-rivera351 Oh boy. Look what we have here. A person who seems to correct every little thing just to feel superior because they want to demonstrate that they know what they're talking about. All you really demonstrate is that you're insecure and probably terrible at that which you boast to practice with such skill. There is something called a multivalued function. This is clearly what I meant by "function." This association is hardly needed when you're talking to people who are somewhat competent. In complex analysis, the definition of function is extended. If you knew that, you wouldn't be trying to correct me. It remains a fact that the nth root is a multivalued function on C, and every nonzero complex number has n different complex nth roots.

@angelmendez-rivera351

2 жыл бұрын

@@PubicGore *Oh boy. Look what we have here. A person who seems to correct every little thing just to feel superior because they want to demonstrate that they know what they're talking about. All you really demonstrate is that you're insecure and probably terrible at that which you boast to practice with such skill.* Being said by someone who is being outwardly and intentionally condescending, and putting their train of insults before any valid arguments, this is just meaningless bluster. Did you get it out of your system? Did that stroke your ego enough? *There is something called a multivalued function. This is clearly what I meant by "function."* I know what it is you are talking about, but your misuse of terminology was worth pointing out. "Multivalued function" is a contradiction in terms. *In complex analysis, the definition of function is extended. If you knew that, you wouldn't be trying to correct me.* No. I know you are wrong, and I have more than sufficient education to know that. Your ignorant self being in a state of denial does not change this fact. I know the study of Riemann sheets is important in complex analysis and projective geometry, and they are also an an important foundation for studying presheafs. It changes absolutely nothing about the fact that you are wrong. No textbook in complex analysis defines function differently than in any other context. If you actually were literate on the subject, you would never talk about "multivalued functions." Almost no one educated on the subject does.

@Gautam-tk8tf

9 ай бұрын

@@PubicGore woah chill, you wrecked that guy-

Find i^i Know: e^(iπ) = -1 Raise both sides to 1/2 th power e^(iπ)^(1/2) = (-1)^(1/2) e^(iπ)^(1/2) = i Raise both sides to the i th power e^(iπ)^(1/2)^(i) = i^i By properties of exponents, we have: e^(i²π1/2) = i^i i² = -1 by definition, in the end we have: e^(-π/2) = i^i

in class i suggested adding "imaginary numbers" to the natraul, integer etc. chart and some student started screaming "A GAJILLION ZILLION!!!" and i did not sleep well that night...

Actually for any root of complex we can use a formula r=√(x^2+y^2) It's modulas Z=x+yi √z=1/√2{√(r+x) +i√(r-x)}

It actually makes sense. If i^1 is a rotation by 90° or π/2 then i^(1/2) or √i is a rotation by 45° or π/4

This was a problem on my first Further Maths test in AS-Levels. Pretty easy but very fun!

Anywhere you see a 1/√2 can be replaced with the following (according to unit circle): cos(π/4) = 1/√2 cos(7π/4) = 1/√2 sin(π/4) = 1/√2 sin(3π/4) = 1/√2 Alternatively you can get negatives: cos(5π/4) = -1/√2 cos(3π/4) = -1/√2 sin(5π/4) = -1/√2 sin(7π/4) = -1/√2 Perhaps these could all be used as substitutes? I tried replacing some of these with 1/√2 in the final equation at the end of the video and I graphed it but the equation looks wildly different. Thoughts?

Imagine a 6th grader watching this video, couldn't be me right?

3 ай бұрын

How's 8th grade going?

@data50090

3 ай бұрын

pretty well so far, math score really good and I am going to be one of the 5 people in my school to do the mathematics national competition in my country (I hope that goes well)

@FlyFlux

2 ай бұрын

​@@data50090Gl

@kaauchpohtato

2 ай бұрын

dang howd that go?

@data50090

2 ай бұрын

@@kaauchpohtato it did not go well

Hello Brian! Your videos are captivating as usual. Really enjoyed it. I was hoping if you could kindly make a video on tensors. It would be really helpful

This is the same as rotating 1/sqrt2 + (1/sqrt 2)i by 45 degrees to get i, which is neat this means the cube root will be 30 degrees. The third root of i is cos(30) + sin(30)i = sqrt3/2 + (1/2)i. The rule is therefore nth root of i = cos(pi/2n) + sin(pi/2n)i

i = e^(i(pi/2 + 2kpi)) => i^(1/2) = e^(i(pi/4 + kpi)) = cos(pi/4 + kpi) + i*sin(pi/4 + kpi) = + or - (sqrt(2)/2 + i*sqrt(2)/2), k being an integer. :)

Also, you can think about sqrt(i) in context of rotation: -1 is turn on 180 degree and if we multiple any number by i twice, it's means we rotate by 90 degree two times and get negative number with respect to original one. So, in that case, we should to find "rotation" of real number by 45 degree. Using knowledge about trigonometry we find out length of imaginary projection and real one the same. So, the answer will be (1/sqrt(2)) * (1 + i)

Calculating it in polar coordinates is much simpler: i is 1 / 90 deg. Consequently sqrt(i) is 1 / 45 deg, or sqrt(0.5)+i sqrt(0.5) (Pytagoras)

sqrt(i) = a+bi has one solution, (a+bi)^2=i has 2 solutions because the square root function is defined as the positive answer to the square root of a number. Also, there is a faster way. 0 = a^2 - b^2 a^2 = b^2 a = b 1 = 2ab 1/2 = b^2 = a^2 a & b = 1/sqrt(2) sqrt(i) = 1/sqrt(2) + 1/sqrt(2)i Just wanted to point that out, also it doesn’t require any powers of 4 so it’s simpler for younger audiences.

@TerryPlays

8 ай бұрын

Hi there.

Some dude that ask: Hey what's your favorite number? Me: √i

@elchile336

2 ай бұрын

that same guy: ...in real values? Me: 4throot(-1)

Just took Complex Analysis last semester. Good vid

You can also consider the complex definition of a "square (complex) number" - root the amplitude and half the angle. With this definition, and the extra knowledge that i has an amplitude of 1 and an angle of π/2 (90°), we can use those two bits of information to conclude that sqrt(i) must have an amplitude of 1 and an angle of π/4 (45°) All we have to do is use the formula for a complex point with amplitude α and angle θ: z=αcos(θ)+αsin(θ)i If we plug in α=1 and θ=π/4 (45°), we get sqrt(i)=1/sqrt(2)+(1/sqrt(2))i

@scrungozeclown836

Жыл бұрын

Of course, on a circle, π/2 radians is the same as -3π/2 radians, and half of that would be -3π/4. Plugging in the values gives you the negative answer. I will say, im not sure why it was surprising to the videomaker that there was two answers? Thats how the square root "function" works for every other number (that fits on the complex plane - i dont exactly know ALL math, so maybe there is a number that has only 1 square root?)

Better insight is to use geometric polar coordinates and makes it easier to see why this works. i is a vector length 1 at 90 degrees. the square root is just same length 1 at (45 degrees) since square root of 1 for the length. Multiplying polar coordinates requires adding the angles. Therefore if I multiply vector 1 at 45 degrees by itself it results in 1 at 45+45 = 90 degrees. Therefore the square root of i is the vector 1 at 45 degrees. This vector in rectangular coordinates is now = Cos(45 deg) + iSin(45deg) = 1/sqr(2) + i/sqr(2) = 1/(sqr(2) (1+i). On the Cartesian plan it is a vector length 1 at 45 degrees. Square root of i. Likewise i^2 = 1(180 degrees) or a vector length 1 at 180 degrees since it is twice 90 degrees. You can take i to any power by just multiplying the exponent by the angle in polar coordinates and then convert to rectangular coordinates. The vector is always a length of 1. The math is easier.

Also, by logical reasoning, if 2ab = 1, then ab = 1/2, and therefore, there are two valid options for the values of a and b. The only known factors of 2 are 1, the Square Root of 2, and 2 itself, as it is prime. Therefore, if it is given that a or b != Square Root of 2, then a = 1 and b = 1/2, or vice versa. Thus, the 2 and the 1/2 cancel out to 1, and 1^2 = 1.

You can also solve for this as: i=cos(π/2+2πn)+i*sin(π/2+2πn) √(i)=cos(π/4+πn)+i*sin(π/4+πn) where n is an integer Since if we have (cos(x)+i*sin(x))^(a)=cos(ax)+i*sin(ax)

2:00 with the left equation we know a = ± b and knowing this the second equation is east to solve

2:00 First way I saw was a and b must be equal so 1/2=a^2 which means a or b=+-1/sqrt2

No, there are not two values for sqrt(i). There is only one. There are two values for z in the equation z^2=i. You get that equation by making the invalid step of squaring both sides of z=sqrt(i). It’s like starting from 1=-1 and then squaring both sides to find that 1=1, and then using that fact to “prove” that 1=-1. Obviously that logic doesn’t work there, so don’t try to use that logic to prove that sqrt(i) has two values. Also, you wouldn’t say that sqrt(1)=-1, would you? But you would obviously say that x=1 and x=-1 in the equation x^2=1.

@user-wu9hy4lt2w

2 жыл бұрын

Your comment is √ i={cos(π/2)+isin(π/2)}^(1/2)=cos(π/4)+isin(π/4)=1/√2 +i1/√2, and -√ i =-{cos(π/2)+isin(π/2)}^(1/2)=-{cos(π/4)+isin(π/4)}=-(1/√2 +i1/√2)=-1/√2 -i1/√2 ? I certainly think it's more reasonable to think so. However, complex numbers that are different from real numbers are difficult for ordinary people to understand. If you have an explanation or material that is easy to understand, please let me know.

2:46 Irrationalizing the denominator

I'm so happy that I managed to get the real and imaginary parts being cosπ/4 and sinπ/4 just by looking at the thumbnail, feels nice to have something make Intuit sense. A neat way to think about this is rotation, multiplication by i gives a 90° or π/2 movement, so √i would be half that

Imagine a unit circle centered on the origin of the complex plane. Start from 1 and rotate a certain angle to get to another number on that circle, for example -1, and to take its square root, take the point that you would get with half that angle. You can convert the numbers with angles by using sin for the imaginary part and cos for the real part. Edit: You can rotate halfway both directions because square roots can be positive and negative.

@angeldude101

2 жыл бұрын

The question is then which direction did you rotate. You can go the long way around the circle, or the short way (-1 had both paths be the same length) the path you took to get the destination determines whether the answer has a negative sign. Half of 360° is 180°, but 360° is the same as 0° and half of 0° is still 0°, and 0° is not the same as 180°.

@marcusscience23

2 жыл бұрын

@@angeldude101 It doesn’t matter because square roots can be positive or negative. sqrt(1) can be +1 or -1.

My solution: knowing that Log(z)= i arg(z) + ln(lzl) (complex log) we can substitute i for z and we get Log(i)=i pi/2 so e^i pi/2 = i now take the square root of both sides so sqrt(i)=sqrt(e^i pi/2) taking square root is the same as putting something to the 1/2 power so we get sqrt(i)=(e^i pi/2)^1/2 now we can multiply the powers (this rule does not always work in the world of complex numbers but it works most of the time) so sqrt(i)=e^i pi/4 now we can a) use euler's formula to evaluate the expresion so sqrt(i)=cos(pi/4) + i sin(pi/4) = sqrt(2)/2 + sqrt(2)/2 i or b) locate the points on the complex plane using the unit circle we know that the point with angle pi/4 and distance from the origin of one has coordinates sqrt(2)/2 , sqrt(2)/2 which on the complex plane will be sqrt(2)/2+sqrt(2)/2 i now we know that l sqrt(i) l = l e^ i pi/4 l wich is always one becouse rel(i pi/4) =0 so we need to multiply everything by one so we get sqrt(i)=sqrt(2)/2+sqrt(2)/2

2:00 that is a massively overcomplicated way of solving it. Just say a^2 = b^2, since 1 = 2ab you know they have the same sign so you know a=b, thus you can just say that a=sqrt(0.5) and you've solved it. Sqrt(i) = sqrt(0.5) + i*sqrt(0.5)

@serulu3490

2 жыл бұрын

Ikr 1=2a² 1/2 = a² a = b= 1/√2

Nicer to see with the "unit circle" (is this the right vocabulary here?) and multiplications of numbers on it as adding the angles

Correction: Imaginary numbers don't have a real part, but complex numbers have both. So it's not confusing anybody.

@SQRTime

2 жыл бұрын

Hi Alex. If interested in math competitions, please consider kzread.info/dash/bejne/fatrpcWFk6a7lZc.html and other videos in the Olympiad playlist. Hope you enjoy 😊

u can also prove it using the fact that e^ix = i, where x = pi/2

It's more simple if you use the exponential form

0:22 correction: all complex* numbers can be written in this form

That's why my teacher doesn't like square roots out of non negative real numbers: since it gives you two different values, it cannot be used as a function

Fun fact: All numbers, including real numbers, can be written in a+bi form a can be anything, but b has to be 0

This turns up in for example micro acoustics where the problem has both diffusion and propagation like properties.

Shouldn’t the result be the “positive result” only because we’re talking about the principal root?

@simong1666

2 жыл бұрын

Think of the case of sqrt(-i)... then the roots are (1-i) and (-1+i) which one qualifies as positive?

@Firefly256

2 жыл бұрын

@@simong1666 When dealing with b⁴ = 1 you only get 1 as answer, not -1 Because sqrt(36) = x x² = 36 Now we have created an extraneous solution x = -6 Because of principal root, sqrt(36) only equals 6. If talking about all roots, sqrt(36) = 6, -6 So because we squared the square root, when we do the inverse of x² = 36, we just do the step x = sqrt(36), and not x = ±sqrt(36) In the video he did ± which I think is wrong if we're talking about principal root, which we usually are

@qwerty687687

2 жыл бұрын

the whole point of complex numbers is to calculate all roots, not just the principal root. As Simon G pointed out, the concept of a principal root doesn't even make sense for roots of complex numbers (with a non-zero imaginary part). So I don't think we atre talking just about principal roots here.

And what do you think about : sqr(i)=j sqr(j)=h ?

Its cool that you used solid math to completely prove something that someone with a small amount of trig skill could have just intuitively deduced. If -1 is a 180 degree rotation on the complex plane and the square root of -1 is i (which is a 90 degree rotation on the complex plane), it stands to reason that the square root of i would be (1/root2) + (1/root2)i. Which is a 45 degree rotation on the complex plane. I wonder if that means that the 4th root of i is a 22.5 degree rotation? And so on. Proofs can be so elegant sometimes. Edit: YEP! The pattern holds! 4th root of i = cos(pi/8) + sin(pi/8)i 8th root of i = cos(pi/16) + sin(pi/16)i and so on. So you could generalize this to: Nth-Root of -1 = cos(pi/n) + sin(pi/n)i Neat!

@WaterMeetsLava

2 жыл бұрын

That's basically De Moivre's theorem

@EvilSandwich

2 жыл бұрын

@@WaterMeetsLava Oh! Good to know! I'll look it up!

you could do this: i = sqrt(-1) sqrt(i)=sqrt(sqrt(-1)) sqrt(sqrt(-1)) = (-1^1/2)^1/2 sqrt(i)=-1^1/4 this is should be an extra complex number because there is no number that can be 4th powered to a result of -1. i^4 = (i^2)^2 = -1^2 = 1

Love this stuff!! Complex numbers are among my favorite mathematical objects!! Thanks lots for posting!! :) :)

@BriTheMathGuy

2 жыл бұрын

Glad you enjoyed it!

You can think of this geometrically by drawing a unit circle on the complex plane.

Thanks! This is amazing!

What a Coincidence! You and Professor Penn posted at the same time

@BriTheMathGuy

2 жыл бұрын

🧐

sqrt(x) is a function of x so it can only have one result, not two.

@mathisnotforthefaintofheart

2 жыл бұрын

Yes, and actually using the square root symbol is abuse of notation when dealing with complex numbers. A better way of putting the problem is "Solve the quadratic equation x^2=i" and that equation has....TWO (complex) solutions of the form a+bi.

@fewwiggle

2 жыл бұрын

In mathematics, a square root of a number x is a number y such that y2 = x So, by definition, you have two solutions

@wiggles7976

2 жыл бұрын

@@mathisnotforthefaintofheart How is it an abuse of notation? You can always just take the root with the smallest angle, even if you have the 5th root of i or any root of i, or any root of any complex number.

@mathisnotforthefaintofheart

2 жыл бұрын

@@wiggles7976 Then that needs to be clearly mentioned in the question. Very often it isn't.

This mightve just been my algebra teacher being picky, but arent you not supposed to leave roots in the denominator? I mean, i understand why 1/sqrt(2) looks better than sqrt(2)/2, but i believe the latter is easier to work with

the way 1^2 = (-1)^2 = 1 made me think that there should be 4 answers to x^2 = i (*) since i is an imaginary number and that other 2 solutions is just 2/4 answer to (*)

Here's a quick solution: Multiply by 2/2 inside square root. √i = √(i x 2/2) = √(2i) / √2 = (1 +i)/√2. Done.

I never liked the definition of a square root, since two different numbers squared yields the same value. We define the square root of -1 as i, but -i works just as well. And how do we decide which square root of an imaginary number is THE square root?

@Anonymous4045

2 жыл бұрын

-i and i arent the same thing tho. One is -1 * sqrt(-1), the other is sqrt(-1) respectively. Yes, when multiplied by each other, they yield -1, but that doesnt mean they are the same

@billmorrigan386

2 жыл бұрын

@John S. The square root is exactly as you logically think it should be. I mean it returns two values, one positive, one negative. For simplification we often use principal roots and because of our laziness (mathematicians can also be lazy) we often don't even bother to mention it (whether we use principal roots or just roots), although folks might have huge problems because of that. Actually, i is defined as +√-1 because √-1 = +i, -i. Nowadays we often define a complex number as an ordered set of two real numbers with a specific rule for multiplication, which means we don't have to deal with complex numbers on a formal level. And that's how it is done in more rigorous math books. Bottom line: there's nothing wrong with definitions. It's just authors neglect finer points and some explanations. *Actually, it's a malpractice in my opinion. All you said, all your critique is 100% valid, while other people just pretend to understand stuff they have no idea about, unfortunately.*

It is not entirely true... the principle root, or square root means the one with the biggest real part. That is, sqrt(4) is always 2 and not -2 even though (-2)^2=4. Using that logic, sqrt(i) can only be the positive one

I think an interesting method to this is realizing e^(i π/2) = i then taking the square roots of both sides e^(i π/4) = i^1/2 From there apply eulers formula.

@andrewkarsten5268

2 жыл бұрын

You miss a solution that way. If you start with e^(2nπi+iπ/2)=i, then take the square root, (to account for periodicity), then you’ll see you actually get two distinct solutions. The other is e^(5πi/4), which when you put into a+bi form gives -√2/2−i√2/2, which indeed gives i when squared.

Another super interesting thing is i raised to the i-th power First the Euler's thing, e^(ix) = cosx + isinx. Plugging in x = π/2, e^(iπ/2) = i. So, i^i = (e^(iπ/2))^i = e^(iπ/2 * i) = e^(-π/2)

2:00 “Pick your favorite way of solving this”. Doesn’t it just include lesser steps to conclude from a^2-b^2=0 that a=+/-b?

Draw a unit circle. Bcoz |sqrti|=1 r=1 do positive 45 degree angle the height is imaginary part the base is real part

If you watch this video and feel it difficult to understand, think about it from the opposite. When "something" becomes squared, it becomes " i ". When squared, it was 1 or -1 to be 1. Once squared, it was i or -i that would be -1. Now, if you think of the real number as a complex number with the imaginary part as 0, that is, 1 =1 +0*i, -1 =-1 +0*i, 1^2=(1+0*i)^2=1^2+(0*i)^2+2*1*(0*i)=1^2=1, (-1)^2=(-1+0*i)^2=(-1)^2+(0*i)^2+2*(-1)*(0*i)=(-1)^2=1. And, if you think of the imaginary number as a complex number with the real part as 0, that is, i =0 +1*i, -i =0+1*(-i), i^2=(0+1*i)^2=0^2+(1*i)^2+2*(0)*(1*i)=(1*i)^2=(i)^2=-1, (-i)^2={0+1*(-i)}^2=0^2+{1*(-i)}^2+2*(0)*{1*(-i)}={1*(-i)}^2=(-i)^2=-1: Likewise, thought of i=(a+bi)^2 in the video method.

I remember back in high school, we had to put the square root of I in the form a + bi. Still makes my brain hurt to this day

When squaring both sides of an equation, be careful not to introduce extraneous roots.

yo I know im prob wrong with my logic since mathemeticians didn't use something even close to as simple as this but what if you think of it as the square root of I = the square root of the square root of -1. Then, put it = x. Square to cancel out the big square root, the square root of -1 = x^2. Then, square that to get -1 = x^4. Then, take the fourth root of both to get i + or - 1 = x. So, the square root of i = i + or - 1. If this somehow does work then ig it doesn't matter cuz its not a real number like your solution

Moivres's formula Z^n = |Z|^n * cis(n*ø) Say that Z = i, then Z^(1/2) = i^(1/2) = k Z also equals to a+bi, so a=0 and b=1 Angle is 90° and |Z|=±1 k = ±cis(90*(1/2) = ±cis(45) = ±[(2^(1/2))(1+i)]/2

The way I found a and b was thus: if 0 = a^2 - b^2, then a and b must be the same. if 1 = 2ab, then ab = 1/2. if a and b are the same and multiply to be 1/2, then each must be the square root of 1/2, AKA 1/sqrt(2).

I'm an Italian high school student and I've just watched your video. I found it amazing! If I were you I wouldn't have a clue how to solve it. Just outstanding!

√2i is superior Because since a+b= √(a²+2ab+b²) If a=1 and b=i It's 1+i= √(1+2i-1) So 1+i= √2i

Another way to look at this; consider imaginary numbers as scaling and rotating numbers on the complex plane. i is at the 90 degree mark. Half a rotation from 1 to i would be a 45 degree rotation, which is at the same number as shown in the video.

3:22 you aren't supposed to keep root in the denominator

Loved it. But it is an odd universe that lets imaginary numbers have a real effect on the physics calculations.

just a rotation on complex axis

Complex analysis is my favorite math branch.

@BriTheMathGuy

2 жыл бұрын

One of mine too!

@magicmulder

2 жыл бұрын

Fun fact: The German term is the misleading “Funktionentheorie”, “theory of functions”. To me it always was too rigid - every continuously differentiable function is infinitely continuously differentiable, where’s the fun in that? Look at all those real function that are differentiable but not even continuously differentiable.

@PXO005

Жыл бұрын

@@magicmulder wait, don't get me wrong I'm just a high schooler but I thought only continuous functions can be differentiable throughout their domain?

All you're doing is changing it from i = sin(90°) to sqrt(i) = +/- sin(45°). Neat.

Me : *watch an math video* Also me : *doesn't understand any single word of what the guy said*

dont reałly need to do much algebra imo. in the same way i is halfway rotated to -1, √i is halfway rotated to i. so in polar coordinates, its 1∠ π/4. but if you want it in normal coords or whatever, itd be (1+i)×√2/2 just going off the unit circle

I mean just use the polar form

@BriTheMathGuy

2 жыл бұрын

also works of course!

finished a complex analysis homework recently and stumbled on this number, this was 7 days ago? dope!

You could've used the exponential form for the answer.

Please increase the volume... It's too low..

Way easier using polar coordinates. i = exp(i.pi/2), thus i^(1/2) = exp(i.pi/4), which is the positive root you found. Of course it's opposite value, -exp(i.pi/4) will also be a valid solution, which yields the negative answer.

It is the 45 degree rotation and at 45 + 180 degree rotation with absolute value 1

what we must be aware is that u(θ) = i^(θ ∙ 2/π) so, for any given angle θ u(θ) returns a complex unit in its angle. so i^2 = -1 and i^(1/2) = (1+i)√2/2 i³ = -i and i ⁻¹ = -i too

It would be better if you also showed it on an Argand diagram and explained how multiplying numbers adds their arguments so the two solutions are halfway around to i in each direction

There’s an easier way to do this. I solved it in a few seconds in my head using polar/exponential form.

Shouldn't the only correct answer be when a and b are both positive? I feel like the reason you get +/- is because you started by squaring both sides, which added another solution

@kchromaticpiano

2 жыл бұрын

You will realize it is not extraneous if you plug it into the original equation though!

@simong1666

2 жыл бұрын

The sqrt(-i) is 1-i and -1+i As can be seen there is no only positive or only negative rule to be followed

There's a more intuitive and geometric way of thinking about it: if a and b are complex numbers on the Argand-Gauss plane, then c = a * b is the vector whose norm is the product of the norms of a and b, and whose argument is the sum of the arguments of a and b. In this case, a = b = sqrt(i) so if r is the the argument o sqrt(i), then r^2 = 1 and hence r = 1. This means that sqrt(i) must lie on the unit circle. So all we need to do now is find the vector on the unit circle such that its argument is half the argument of i(which is pi/2). That gives us c = cos(pi/4) + sin(pi/4)i, the "positive" answer. If course if c^2 = i, then so does (-c)^2 and we're done.

@SQRTime

2 жыл бұрын

Hi Cabutchei. Thanks for sharing your thoughts. If interested in math competitions, please consider kzread.info/dash/bejne/fatrpcWFk6a7lZc.html and other videos in the Olympiad playlist. Hope you enjoy 😊

Cool , another way to do it is by the eulers identity, e^i(a)= cos(a)+i*sin(a), if we replace a=pi/2 then cos=0 and sin=1 so e^i(pi/2)=i and we can extract the square root (elevate to 1/2) and by properties of the exponents we have that square root(i)=e^i(pi/4) sorry for my english, i'm learning english and math with this Chanel and is cool

@BriTheMathGuy

2 жыл бұрын

Nice! Thanks and thanks for watching!

Imaginary numbers are not complex numbers. Both real and imaginary numbers are 1 dimensional numbers representing a point on either the real or imaginary number lines. A complex number is 2 dimensional representing a point on the complex plane.

@user-wu9hy4lt2w

2 жыл бұрын

Your comment "Imaginary numbers are not complex numbers" is strange. If the imaginary number is not a complex number (as is the case with real numbers), it should not be represented in a complex plane. The complex number was mentioned in the dictionaries as follows: Encyclopedia MyPedia "Real x,y and the number that can be expressed as z=x+yi by imaginary units. y=0 is a real number, x=0 is a pure imaginary number. ・・". Britannica International Encyclopedia "Any complex number z is written in the form of z=a+bi with a and b as a real number, in this case, if b=0, z is real a, so the range of complex numbers includes real numbers. ・・". In other words, if you do not think about the complex number even if the real part and the imaginary part are 0, the explanation of this video will not be possible.

@angeldude101

2 жыл бұрын

Every real number is a complex number with the imaginary part equal to 0. Every imaginary number is a complex number with the real pay equal to 0. Every integer is a rational number with the denominator equal to 1.

2^(-1/2)+i 2^(-1/2) and -2^(-1/2)-i 2^(-1/2)

@MitchBurns

7 ай бұрын

It’s east if done in polar. i is just 1 at 90 degrees. So it’s square root is 1 at 45 degrees as well as 1 at (45+(360/2)) degrees.

Sorry to correct you, but your first mentioned statement is wrong. i is not defined by i=V(-1) If i would be defined this way, you could calculate i² by the following way: i²=(V(-1))²=V(-1)*V(-1)=V(-1*-1)=V(1)=1 which obviousely is incorrect. Therefore i is defined by i²=-1

using the properties of square roots, and the fact that i=sqrt(-1), you can assume that sqrt(i) would just be the fourth root of -1, which is also an imaginary number, therefore the most simplified way of writing it is actually just how you wrote it, which is as sqrt(i)

Square Root of NEGATIVE (Square Root of -1).

I'm kind of proud I did it myself

Very useful video, I had previously thought that sqrt(i)=-i.

@Ostup_Burtik

6 ай бұрын

No, because (-i)²=-1 like i²=-1

@meeptothemax375

6 ай бұрын

@@Ostup_Burtik Which I realized after watching the video.