A Groundbreaking Mathematical Result

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Thanks Leibniz, very cool! Without complex numbers or the notion of negative arguments in roots, Leibniz shook the mathematical community back in the days by denesting radicals! We are going to explore his result for sqrt(1+sqrt(-3))+sqrt(1-sqrt(-3)) giving us sqrt(6). As a corrolary using the conjugate and difference of two squares, we also find out the value for sqrt(1+sqrt(-3))-sqrt(1-sqrt(-3)) in the process. Enjoy! =)

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Postfach 11 15
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[PapaFlammy]

*_Thanks for watching, I hope the video was to your liking <3 Make sure to provide the community with some sqrt() multiplication insights if you can! :)_* Apart from that, here are all of today<'s relevant links

[luggepytt]

I would like to address the question that Jens was kind of begging the audience for an answer: When is it valid to say that √(z1) × √(z2) = √(z1 × z2) if z1 and z2 are complex numbers?

First, we need to clarify what we really mean when we write √(z).
We want the expression √(z) to have a single, well-defined value, but the equation w² = z has two solutions (unless z = 0). So which one shall we choose?

The common practice is to choose the one with a positive real part. Or, when the real part is zero, the one with a positive imaginary part.

This is called the principal square root of z. (And the definition works when z is real too.)

Some examples, following this definition:
√(4) = 2 (and not -2)
√(2i) = 1+i (and not -1-i)
√(-2i) = 1-i (and not -1+i)
√(-4) = 2i (and not -2i)

An alternative, but equivalent, definition of the principal square root is this:
If you write z as r × e^(i𝜋ϑ), where r > 0 and -𝜋 < ϑ ≤ 𝜋
then √(z) = √(r) × e^(i𝜋ϑ/2)

If we follow this definition, then it is easy to see that the equality
√(z1) × √(z2) = √(z1 × z2)
holds under the following circumstances:

Suppose z1 = r1 × e^(i𝜋ϑ1) and z2 = r2 × e^(i𝜋ϑ2)
Then everything is fine as long as -𝜋 < (ϑ1 + ϑ2) ≤ 𝜋

In the special case with two complex conjugates with non-zero imaginary parts, as we had in this video, ϑ1 + ϑ2 = 0, so the equality holds.

So Jens' calculations were flawless. Hooray!

[tiagolevicardoso]

Hill: question mark is a valid variable.
Proof: Let ? be a variable
Q.E.D

[YourPhysicsSimulator]

"Mathematicians are serious, boring and quiet people"
What mathematicians trully are doing: 0:09

[redhotdogs]

you can also do this by writing 1+sqrt(-3) as 2exp(pi*i/3) and 1-sqrt(-3) as 2exp(-pi*i/3)

this then simplifies into

which equals
sqrt(2)(2cos(pi/6))
which is
sqrt(6)

[Brien]

these were roots of the polynomial I was given in my abstract algebra class. We were asked to determine the galois group and all the indermediate field extensions. In general are polynomials of degree 4 with these kinds of roots very interesting! Everyone likes the D4 group after all .

[Vaaaaadim]

Sneezed for his intro, what a time to be alive

[parasb]

You know what i guess the catch here is changing that question mark to x . I solved i similar question 2 years back was stuck for like two days until i realised "X" is the saviour !!

[realkabirc]

This Problem was really easy, I didn't go into the algebra, I just converted in the Polar form and then simplified and got root 6.

[MrRyanroberson]

oh man you got so close to something even cooler. now that you have x and y, you can take x+y and x-y to find algebraic simplifications for sqrt(1±sqrt(-3)) to be 2(sqrt(6)±sqrt(-2))

[renanrodz]

Someone: "Do you have plans for this weekend?"
Me: "It's complex..."

[brandonklein]

I mean if we want to be very careful, we can take the square roots as complex numbers and assure that the complex parts cancel. I would guess that if we did this and went through, we would find 2 more solutions due to the +- on √z for complex z.

[hardikjoshi]

Loved your DJ flammy version in the last video.. Today I adjusted my neck at the angle of 45 degrees so to see my gf showing her nested radicals.
Thanks sir you did a great help solving the problem. Now I can enjoy my remaining day and will share this story with Leibnitz😂
❤️
🇮🇳💟🇩🇪
Your boi always:)

[pardeepgarg]

Alt title : A Math breaking Ground Result

[johnny_eth]

The numbers inside the radicals are complex, so you need to work with both roots. If you work with the polar form, you reach two possible values ±sqrt(6) and ±sqrt(2).i.
By squaring everything in the start, you are basically merging both roots into its squared value and hence get one answer.

[XAE-ycrr]

> 5:00 AM
>slides into view like a boss
>"Gyee...*coughs*! "

This is already a great start of the day.

[ricardoparada]

This reminds me of a similar process when u use the rational roots theorem to find out if a number is rational or not. It’s really nice

[Mrballen]

Hey flammy, theres also an approach via a std method called square root of complex no where u assume each term as x+ or -iy, simply sqare, find x and y, then basic addition.

[Mike-frhp]

if x equals to sqrt(1+sqrt(-3))+ sqrt(1+sqrt(-3)) but at the same time x equals to sqrt(6), sqrt(1+sqrt(-3))+ sqrt(1+sqrt(-3))=sqrt(6)

[tambuwalmathsclass]

I didn't try this, but I believe if I were to simplify it, I'll end up with a complex Solution