OMG, It Works! Solving a Mystery with Euler's Identity (MindSphere shorts)#maths

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20. Trendy How to solve equations with Euler's identity
Euler's identity explained simply
Beginner's guide to Euler's formula
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[xxAnonymous_Justicexx]

Well -1 cannot have a squareroot because it is negative and a square of a number is always positive because when two numbers are multiplied to get a square they have to be of same sign and same value to get a square and when two signs are same and multiplied we get positive. In this equation i thought in another way where we take
=> -1^((1/2)×(-1)½)
=> -1^((1/4)½×(-1)½)
=> -1^((-1/4)½)
=> -1^((root -1/2)(Using basic rules i got till this much)
OR
Let root -1 be i because,
(Root -n = ni)
And n is 1 and -1 is i
We get,
=> -1^(i/2)
So in terms of i, we got -1^(i/2).
Now we will take root -1^i
Since root -1= i
We get,
i^i...
There is no precise calculation....though.

I just came in 10th so i didn't learn logarithms or calculus fully yet. But I tried to use basic things I know to get the answer. Is it correct?


Now,
-1^(root-1/2)= e^(-pi/2)
But isnt root -1 = i? Then how come pi is taken?.

[koopa]

pi managing to sneak into every math equation fr

[LiminalUniverse]

The solution is incorrect even though the answer is correct.
Since, (a^(b))^(c)=a^(bc) is only true for real numbers b and c.
A more careful analysis well lead to a correct solution.
Here is the solution,

i^i = e^(ln(i^i))
We can substitute e^ln since they are inverse functions they cancel each other out.
Using properties of logarithms.
ln(i^i)=iln(i)
Substitute.
e^(iln(i))
The natural log of a complex number can be represented as ln(z)=ln(|z|)+iarg(z)
|z|=sqrt(x²+y²) where x is the real part and y is the imaginary part.
We can represent i as z=0+1i where 0 is the real part and 1 is the imaginary part.
Substitute that in.
|i|=sqrt(0²+1²)
=1
And natural log of 1 is always zero.
ln(1)=0
ln(i)=iarg(i)
arg(i) represent the argument of i or the angle on the complex plane.
arg(i)=(π/2)+2πk
Where k is any integers.
(π/2) is the angle in the complex plane at which i is located.
The 2πk tells us that we can rotate another 2π at the same point.
We well go for the principle value where k=0.
Now we have,
ln(i)=i(π/2)
Substitute that in e^(iln(i))
We get,
e^(i×iπ/2)
i×i=i²=-1
We finally get,
e^(-π/2)
The same answer with the correct solution.

Edit:Complex Logrithms do have the same property as normal logrithms.
Complex logrithms needs to have the angle theta between (-π, π).
ln(z)=ln|z|+iarg(z)
Where, arg(z)=θ+2πk
Where k is any integer
And θ is in the interval (-π, π)

[talharuzgarakkus]

i^i doesn't have just one answer you must show e^(-2pi*n-pi/2) (n E Z)
I suggest to use euler transformation for the imaginary powers.

[element]

This isn't the most rigorous method to use, even though it gets the right answer. We have to be a little bit more careful when exponentiating complex numbers.

Remember that z^w = e^(wlnz) = e^w · e^(lnz). Hence, all complex exponentiations where z≠0 can be converted to base e. Now, how can we write a logarithm of a complex number in terms of real numbers?

Recall that for any complex number z and integer k, z = re^(i(θ+2πk)). Taking the logarithm of that, we get ln(re^(i(θ+2πk))) = lnr + i(θ+2πk).

Since r must be a real number, lnr by convention has only one value. r is the radius in polar coordinates, which is the distance from 0 to z, which is |z|. θ+2πk is the angle from the real axis to z, which is arg(z).

Therefore, lnz = ln(re^(i(θ+2πk))) = ln|z| + iarg(z). Since k can still be any integer, arg(z) has infinitely many values, the principal of which is when k=0.

Now, remembering the beginning, z^w = e^(w(ln|z| +iarg(z))). If we let z = w = i, we get i^i = e^(i(ln|i|+iarg(i))). |i| is 1, and ln1 = 0. arg(i) is π/2+2πk, because i is 90° from the real axis.

Now we have i^i = e^(i(iπ/2+i2πk)). Distributing gets us e^(-π/2+2πk), and when k=0, that's ~0.20788

[ankitchauhan]

When I am studying this chapter same doubt arises in my mind
But my teacher said this is not in your syllabus

[Pookiehuji]

Studied in Complex numbers
Make a video on Rotation of complex numbers also

[elegantblue]

Btw it can have many more values other than e^-pi/2

[ankurpareek]

I love the face that (imaginary)^(imaginary) is a real number !

[mathssolverpoint]

This is a particular solution not general solution

[dwijendraprasaddas]

That's why complex analysis is vary important

[youraverageinternetuser]

Also don’t forget that pi^pi^pi^pi COULD equal an integer for all we know

[happyman_smiling]

The rules of real number addition subtraction multiplication do not apply for complex numbers only a post graduate person can do that, so everytime you multiply divide add subtract complex numbers you won't be correct all the time

[Zwischenzug]

That is why, always remember that your imagination can always be a reality if you put your efforts into doing it. MOTIVATION RAHHHH!!

[netanelkomm]

The significance of the constants pi and e amazes me each time.

[Very.Crazy.Math.Pistols]

Is that the prototype of all transcendents or algebraics equal to a^b, with a, b, algebraic number?. I wonder if algebraic^algebraic covered all the reals, because they are dense on the line, every interval has at least one. But they are still countable, and there are still some... 👀

[firefoxclub]

∞ <\>\= ∞ ( Infinity is greater or lower or equal to another infinity)

[yurajek]

bro solved the unsolvable question he made himself

[itz_mario.]

Answer is right but step is wrong, you cant multiply the upper power by lower power when they include "i", i saw bprp do it the correct way

[PrajwalV]

It's amazing 🎉🎉🎉
But i already knew it
But one thing i should say that
You are making videos about beautiful things in maths
But why you never made a video on

Golden ration and silver ratio of mathematics i think it will be best
❤❤❤❤❤❤❤❤❤❤❤❤❤❤