A Quick and Easy 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!!! 🧡🥰🎉🥳🧡
/ @sybermathshorts
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/...
Follow me → / sybermath
Subscribe → www.youtube.co...
Subscribe → www.youtube.co... (shorts)
⭐ Suggest → forms.gle/A5bG...
If you need to post a picture of your solution or idea:
in...
iˣ = 1, i^x=1
#ChallengingMathProblems #ExponentialEquations #Exponentials #ComplexNumbers
via @UA-cam @Apple @Desmos @NotabilityApp @googledocs @canva
An Interesting Exponential Equation with x^x^x: • An Interesting Exponen...
Solving A Challenging Exponential Equation: • Solving A Challenging ...
PLAYLISTS 🎵 :
Number Theory Problems: • Number Theory Problems
Challenging Math Problems: • Challenging Math Problems
Trigonometry Problems: • Trigonometry Problems
Diophantine Equations and Systems: • Diophantine Equations ...
Calculus: • Calculus

[@moeberry8226]

It’s not just multiples of 4, the general form is 4n/(4K+1) where n and k are integers.

[@musicsubicandcebu1774]

Why?

[@moeberry8226]

@@musicsubicandcebu1774 because Syber only gave the base case of i, it can be written as e^i(pi/2 +2kpi). Where k is an integer and can be different from n. You can try it and see that it works.

[@Qermaq]

If n and k are 1, this implies 1^(4/5) = 1. While WA does give this an an answer, the top result is cos(2pi/5) + 1sin(2pi/5). I wonder why the real result is buried here.

[@moeberry8226]

@@Qermaq because Wolfram Alpha gives all complex roots of 1.

[@SyberMath]

My understanding based on what @usdescartes said in a comment:
i^(4n)=1 so i^4=1 for n=1
However, i^(4/5)=(i^4)^(1/5)=1^(1/5) and this is multi-valued...one of its values is 1 but it's not always 1.
I'm new to this so I am not saying this with utmost confidence

[@gheffz]

Well done. Great explanation. Very clear.

[@SyberMath]

Glad it was helpful!

[@RexxSchneider]

But i = i * 1, so the general polar representation of i is e^(2mπi + πi/2), making i^x = e^((2mπi + πi/2)x). We set that equal to e^(2nπi), which is 1. Taking natural logs, we have:
(2mπi + πi/2)x = 2nπi. That gives the general form x = 2nπi / (2mπi + πi/2) which simplifies to x = 4n / (4m + 1). Now the question arises whether the m and n are independent.
If m = 0, then n can be any integer as shown in the video, since i^x = e^(πi/2*4n) = e^(2nπi) = 1 for all n ∈ ℤ.
If m = 1, then x = 4n/5 and i^x becomes e^((2πi + πi/2) * 4n/5) = e^(8nπi/5 + 2nπi/5) = e^(10nπi/5) = e^(2nπi) = 1 for all n ∈ ℤ.
if m = 2, then x = 4n/9 and i^x = e^((4πi + πi/2) * 4n/9) = e^(16nπi/9 + 2πi/9) = e^(2nπi) = 1 for all n ∈ ℤ. And so on.
In other words, if we choose to write i as i^(4m+1) -- which we always can -- then we find that the general solution to i^(4m+1)x = 1 will be of the form x = 4n/(4m+1).

[@BlaqRaq]

Fallacy. Recheck your work. The basis is false.

[@horaciovillegasarango3278]

Muchas gracias Profesor!

[@mitri4939]

Another fun problem and another great video! May I ask where you got your education in Mathematics?

[@SyberMath]

TR. Thanks!

[@goldfing5898]

x = 4 is a solution, because i^4 = (i^2)^2 = (-1)^2 = 1.
Also, multiples of four. Let n = 4k, then i^n = i^(4*k) = (i^4)^k = 1^k = 1.
Does this also hold for k = 0, -1, -2,... ?

Yes, because the absolute value doesn't change for 1 and 1^(-1) = 1/1 = 1.

[@vaibhavjoshi5018]

Please solve JEE Advance and maths Olympiad tough questions for application and concepts building.

[@mathswan1607]

x=4n

[@vladimirrodriguez6382]

Maybe for more generality of the solution, i=e^(pi/2+2kpi), k in Z. So the solution is:
x=2n/(1/2+2k)=n/(1/4+k), n and k in Z.

[@BlaqRaq]

But, if I=sqrt (-1) and (-1)^2=1 then it strictly follows that i^4n ,such that n is integer including zero, would equal 1.
And all odd powers are imaginary and all 2n ≠ 4m, for all n,m members of integers, would be negative.
So, we are sure, we capture all values of x for i^x = 1.

[@kianmath71]

X = 4k, in which k is an integer, great video as always

[@rakenzarnsworld2]

x = 4n (n is integer)

[@bertrandviollet8293]

I don't understand, i should also be written as e^i(pi/2 +2n×pi)

[@giuseppemalaguti435]

4,8...

[@mircoceccarelli6689]

i^x = 1 x = ?
x = 4 n , n € Z . n numero intero !!!!!
😊

[@user-pp2xc6ky4c]

смешно. что то там считает. на комплексной плоскости i это 90, i^2 =-1, следовательно нам надо ещё столько же. то есть i^4=1.

[@Packerfan130]

i^x = 1
i = e^(pi/2 + 2pi N)i
1 = e^(2pi M)i where N and M are integers
i^x = 1
e^(pi/2 + 2pi N)ix = e^(2pi M)i
(pi/2 + 2pi N)ix = (2pi M)i
((1/2) + 2N)x = 2M
(4N + 1)x/2 = 2M
x = 4M/(4N+1)

[@jonnoring7225]

Hmmm, I get x = 2m/(1+4n), where m and n are any integers. Will have to redo my work.

[@Packerfan130]

you're half right since x = 4m/(1+4n) is correct

[@taxttime6216]

I am really wondering if the equation was
i^x=-1
What is the value of x in the form of n
As i^x=1 (x =4n)

[@taxttime6216]

I got the answer

[@DeJay7]

x = 4n + 2, for natural numbers n

[@RexxSchneider]

In simplest terms, we can see that we can set e^((πi/2)x) = e^(2nπi + πi), because e^(2nπi + πi) = -1 where n ∈ ℤ.
So x = (2nπi + πi) / (πi/2) = 4n+2.
However, we can write i as i^5 = i^9, etc. In general i = i^(4m+1) where m ∈ ℤ.
So depending on how we write i, we will find some more solutions of the form x = (4n+2)/(4m+1), which are the solutions of i^(4m+1)^x = -1.

[@rubinkatz9850]

weird, ad for a BNW model iX

[@user-pb2sx9xq5g]

x=n/4; x=4n 😁

[@SyberMath]

😄

[@ertugrulberatbilici]

X=4n

[@alextang4688]

use polar form???🤔🤔🤔🤔🤔🤔

[@loblud4437]

i*i*i*i= 1

[@monkeblazer3154]

bruh x = 4 or 0

[@yttyw8531]

Or 8 or 12...

[@bunnygod3948]

The general solution is 4n

[@monkeblazer3154]

@@yttyw8531 yea yea multiples of 4 are valid sorry forgot to mention and thanks for mentioning

[@Packerfan130]

@@bunnygod3948 the general solution is 4n/(4m+1)

[@epsilonxyzt]

Too much talk.