How To Solve A Complex Exponential Equation

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[@williamperez-hernandez3968]

The sqrt(2 Pi n) was not distributed correctly. The arguments of cosine and sine are sqrt(Pi n).

[@bernarddoherty4014]

15 yard penalty. Second down. (Don’t worry! We always know you come back and score a touchdown!! )🤣🤣🤣😜😜

[@TheRhythmOfMathematics]

Once I composed this problem, I had not thought about complex solutions. My domain of variable is limited in real numbers.
Thanks for discovering the complex roots.

[@bernarddoherty4014]

15 yard penalty. Second down. 🤣🤣🤣. Don’t worry! We always know you’ll score a touchdown! 🤪🤪

[@bernarddoherty4014]

Above was not meant for this entry.

[@XJWill1]

Your solution method was flawed since you cannot manipulate exponents and log functions involving complex numbers using identities that are only identities for positive real numbers. It is concerning that you persist in these mistakes despite having a channel about complex numbers.
Also, you need to define the problem better. The usual convention is that ln(x) is only defined for positive real values of x. For complex analysis, you should use log(z) or Log(z) where that is the complex natural logarithm function. The lowercase log(z) is mult-valued, while the uppercase Log(z) is the principal branch of the function.
If we assume that the problem you meant to solve is actually:
z^(Log(z)) = 1
where Log(z) is the single-valued Principal branch of the complex natural logarithm function, then there are 13 values of z that satisfy this equation. 1 real value, and 12 complex values, which can be written as
z = exp( (-1)^m * sqrt(k*pi) * (1 + i * (-1)^j ) ) for m = {0, 1} , j = {0 , 1} , k = {0, 1, 2, 3}
Furthermore, you did not show all of the solutions from W. Alpha. You need to keep clicking on "more roots" to show them all.

[@bernarddoherty4014]

They’re not called ‘complex solutions’ for nothing! Lol.

[@KennethChile]

Could you explain step by step how did you get this: "z = exp( (-1)^m * sqrt(k*pi) * (1 + i * (-1)^j ) for m = {0, 1} , j = {0 , 1} , k = {0, 1, 2, 3}"

[@XJWill1]

@@KennethChile I'll just give some guidelines. Instead of using identities that are only identities for positive real values, these types of problems should be solved using the definition of exponentiation and the properties of the complex natural logarithm function.
z^w = exp(w * Log(z)) definition of exponentiation
Log(z) = ln|z| + i * Arg(z) where -pi < Arg(z)

[@KennethChile]

@@XJWill1Thanks for the guidelines. I will study them with paper and pencil now.

[@TheRhythmOfMathematics]

@@XJWill1 your explanation is detailed and easy to understand. Thanks for that.

[@ColoradoSockies]

Where’s my lambert W function? :)

[@tamilselvanrascal5956]

🎉

[@user-rn3hd6hc4f]

at 6:48 error

[@KennethChile]

It's correct. sqrt(2)/2 = 1/sqrt(2)

[@maxvangulik1988]

if a^b=1, at least one of the following must be true:
•a=1, b€R
•a≠0, b=0
•a=-1, b€E
case 1: x=1, lnx=0
case 2: lnx=0, x=1
case 3: x=-1, lnx=ipi(2k+1)
(-1)^i=e^-pi(2n+1)
1=e^(-ipi^2(2k+1)(2n+1))
the product of 2 odd integers is another odd integer, so one factor can go away
1=e^(-ipi^2(2k+1))
-pi(2k+1)=2n
if there exist an even and an odd integer whose ratio is pi, then that would be a solution. A common approximation for pi is 22/7, but that's only to 3 digits (if you round pi)

[@maxvangulik1988]

identity: x=e^lnx
e^ln^2(x)=1
ln^2(x)=0
x=1

[@rakenzarnsworld2]

x = 1

[@giuseppemalaguti435]

(lnx)^2=0...lnx=0...x=1

[@chandrikabiswas5829]

X= d/dx(x) + 2-2 🥱🥱

[@lucaswarnke3668]

x^ln x=1
x=(ln x)th root of 1
x=1

[@Feministsiken3169]

Bi halt anlamadım