A Very Interesting Exponential Equation | iˣ = -1

🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts).
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡

If you need to post a picture of your solution or idea:
i^x=-1 iˣ = -1
#ChallengingMathProblems #ExponentialEquations #ComplexNumbers
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

PLAYLISTS 🎵 :

[MisterPenguin]

Does that negative sign look really big to anyone else?

[XJWill]

I think you intended this to clarify and teach about how to properly interpret and manipulate complex numbers to solve algebraic equations. However, I suspect this will actually cause more harm than help. There are basically two issues here that need to be addressed.

This first one is simply addressed. It is how to interpret complex exponentiation such as
z^w
The point that needs to be made is that technically, this is a multi-valued function. There are an infinite number of possible values. HOWEVER, and this is a big however, in the vast majority of cases, people assume that z^w is single-valued, and equal to the so-called "principal value" that comes from choosing the principal branch of the complex logarithm function. Note that Wolfram Alpha makes this assumption by default. It assumes z^w is single-valued and chooses the principal value.


The next point is not so simply addressed, but it is critical, because it is the proper way to solve equations with complex variables. Your "method" of writing constant real values as multi-valued with exp(i*n*2*pi) is not the correct way to do it, and you really should stop that, and switch to the proper method (because your "method" will not work in the general case, and the correct method will).

The standard way mathematicians deal with z^w in complex analysis is to rewrite it in terms of the multi-valued complex logarithm function, which is usually written as log(z), but for clarity I will write it as lc(z) in this comment.
We write:
z^w = exp(w * lc(z)) by DEFINITION
Note that complex analysis starts by defining the exponential function exp(z), which is single-valued, and the complex logarithm function log(z) or lc(z), which is multi-valued. Then z^w is defined in terms of those functions.
Now the important point is how to deal with lc(z). Here is the proper method.
lc(z) = lc(r*exp(i*t)) = ln(r) + i*t + i*n*2*pi
where ln(r) is the real single-valued natural logarithm function, n is an integer index, and z has been rewritten in polar form as
z = r*exp(i*t)
where r and t are real-valued.
That is almost all you need to handle most equations with complex variables. Just write the complex-valued variables in polar form (or sometimes rectangular form, depending on the equation) with real-valued variables. Then rewrite any exponentiation using exp() and lc(), and finally rewrite each lc(z) as a multi-valued function with an independent integer index for each lc() function. Of course, solving the equation is another thing entirely. You need to separate each equation into real part and imaginary part and equate those, so each equation turns into two equations. And quite often, they can only be solved numerically in all but the simplest problems.

[djgiesz]

You can write i^x as (e^(pi/2+2*k*pi)*i)^x, which is the same as e^(x*(pi/2+2*k*pi)*i), and you can write -1 as e^((pi+2*n*pi)*i). Since the base is the same you only have to compare the exponents which means and dividing by (pi/2+2*k*pi)*i gives us which is the same as x=(pi+2*n*pi)/(pi/2+2*k*pi) where "k" and "n" are integers

[GreenMeansGOF]

Still not sure why we don’t use two integers

[lawrencejelsma]

A complex number has no finite solution for x = 4n + 2. This expression lists an infinite amount of solutions basically repeating the same information. It is better to find the first x = 2 (at n = 0) so that we define a single solution from the domain 0 < the Complex Euler Circle angle <= x (or x = 2) times 2π new range formed to not consider repeated information of x in { +/- 2, +/- 6, +/- 10, ...} an infinite set that represents a cosine and sine function same result. Notice also we have to choose unique subset {(0, complex angle ∆, 2π]} domain or subset {[0, complex angle ∆, 2π)} domain in order to have a unique solution where the equivalents in the other domains confuse unique solutions. In the unique x = 2 first n = 0 case the range became the subset {(0, the new complex angle result, 2πx (or evaluated 4π)]} or subset {[0, the new complex angle result, 2πx or 4π)} range formed. This is to avoid repeated infinite other solutions that were never on the initial 0 to 2π domain visualized Complex Euler Circle when we plotted angle of i from the (1)(Cos(π/2) + iSin(π/2)) and the -1 from (1)(Cos(π) + iSin(π)) we were equating for the domain of the Euler Circle. ... The second part of the video playing around for meaningless and erroneous results of an infinite set going into another different infinite set (since n and k assumed unique integers; and, not with the produced error of these infinite solutions: one particular case example of x = 2/5 in that illustrated play now erroneously showing i^2/5 = -1 becomes false result). i^x is best considered how it divides into -1 on the Complex Euler Circle for hopefully one unique solution (notice at x=2 or π range element and the 3π range possible element ignored because that happened from π/2 domain at x = 6 equivalent not considered for a unique solution).

[goldfing]

We have a cycle of four instances:
i^1 = i
i^2 = i * i = -1
i^3 = -1 * i = -i
i^4 = -i * i = -(i^2) = -(-1) = 1
i^5 = 1 * i = i
i^6 = i * i = i^2 = -1
...
So we have a repeating four cycle i, -1, -i, 1.
And i^x = -1 has the integer solutions x = 2, 6, 10, ...

I general, x = 4k + 2 = 2*(2k + 1), for k = 0, +-1, +-2... any integer k.
Proof:
i^(4k + 2)
= i^(2*(2k + 1))
= (i^2)^(2k + 1)
= (-1)^(2k + 1)
Here you can already see that it is (-1)^(odd number), thus -1.
= (-1)^(2*k) * (-1)^1
= ((-1)^2)^k * (-1)
= (1)^k * (-1)
= 1 * (-1)
= -1

[mircoceccarelli]

i^x = - 1 --> x = 2 + 4n,
n numero naturale !

i^2 = - 1 , i^4n = 1 , n numero naturale!

[thomaslangbein]

The positive sqr of -1 is +i. -i is the negative sqr of -1

[MathsScienceandHinduism]

If I put i = root -1 then x/2=any odd integer so x=2(2n+1); where n is integer