This Is My New Favorite Number

i is the imaginary (complex) unit, but what is its square root?

We'll use some complex number theory and matching coefficients to solve a system of equations to figure this one out!

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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

[BriTheMathGuy]

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[riginator]

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

[kingbeauregard]

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

[jeff-]

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

[hassan]

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

[_Obi]

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

[userhomer]

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

[BriTheMathGuy]

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).

[PunmasterSTP]

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!

[pNsB]

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

[TamjidANoor-shbv]

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)}

[JakubS]

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

[reevsy]

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)

[maximilianarold]

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

[RyanthePokemonTrainer]

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.

[theuserings]

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

[KazuyaMLBB]

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

[tcthebeast]

Just took Complex Analysis last semester. Good vid

[charliebowditch]

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

[matejnovosad]

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