💾 Archived View for gemini.locrian.zone › library › stackexchange › jtothej.gmi captured on 2023-12-28 at 16:16:25. Gemini links have been rewritten to link to archived content
⬅️ Previous capture (2023-09-28)
-=-=-=-=-=-=-
Original post by user503220 on math.stackexchange
First of all, I am do math only for fun, and I am only amateur, so excuse me because of my lack on knowledge in this field. I was curious about some answers and I want to know if my judgment was ok or wrong. First I wanted to find out the answer of:
i^j
(where i^2=−1,j^2=1,j≠−1,j≠1), j is split complex number
j^j j^i
Maybe I was wrong to combine complex number with split numbers. I don’t know. Please let me know If my calculus is ok or there are some mistakes in it.
Thank you.
First:
ln(j)
I use the formula for logarithm of split complex numbers:
ln(a+bj)=½(ln((a+b)(a−b))+½j(ln(a+b)/(a−b))
I know that (a+b)(a−b)<0, we will get a complex number in our answer and that that the split complex logarithm is not closed. Of course for all formulas I take into consideration only the principal branch.
So:
Ln(j)=½((ln(0+1)(0−1))+½j(ln(0+1)/(0−1)) =½ln(−1)+½jln(−1) =½(iπ)+½j(iπ)
BUT, when I wanted to verify I am correct I discover that, the answer ½(iπ)+½j(iπ)is actually ln(−j).
I also discover that are at least two correct answer for ln(j)
and for ln(−j).
Ex. for ln(j):
Ln(j)=iπ/2–jiπ/2
and
Ln(j)=−iπ/2+jiπ/2
Proof:
e^(iπ/2–jiπ/2) =(e^(iπ/2))∗(cosh(−iπ/2)+jsinh(−iπ/2)) =(e^(iπ/2))(0−ij) =(−ij)e^(iπ/2) =(−ij)(cos(π/2)+isin(π/2) =−iji=j
I hope is nothing wrong in my calculus.
So
j^j=e^(jln(j)) =
1)
e^((iπ/2–jiπ/2)j)=e^(jiπ/2–iπ/2)=−j =e^(−iπ/2)(cosh(iπ/2)+jsinh(iπ/2)) =e^(−iπ/2)(0−ij) =−ij(cos(−π/2)+isin(−π/2)) =−ij(0−1i) =−j
2)
e^(−iπ/2+jiπ/2)j =e^(−jiπ/2+iπ/2) =−j
because
Ln(j)=iπ/2–jiπ/2
and
Ln(j)=−iπ/2+jiπ/2 =e^(iπ/2)(cosh(−iπ/2)+jsinh(−iπ/2)) =e^(iπ/2)(0+ij) =ij(cos(π/2)+isin(π/2)) =ij(0+1i) =−j
So...
j^j=−j
.... Am I wright?
Ln(i)=iπ/2 j^i=e^(ilnj)
1)
e^(iπ/2–jiπ/2)i =e^(−π/2+jpi/2) =(e^(−π/2))(cosh(π/2)+jsinh(π/2)) =0.521606...+j0.478393...
2)
e^(−iπ/2+jiπ/2)i =e^(iπ/2−iπ/2) =(e^(π/2))(cosh(−π/2)+jsinh(−π/2)) =12.070346...−j11.070346
Am I wright?
i^j =e^(jiπ/2) =cosh(iπ/2)+j(sinh(iπ/2) =0+j∗i
So ... i^j=ij
.... Am I right?
Proof:
i^j=ij jlni=ln(ij) ln(ij)=½(ln(0+i)(0−i))+½j((ln(0+i)/(0−i)) ln(ij)=½ln1+½jln(−1) ln(ij)=(ipi/2)j jlni=ln(ij) j(ipi/2)=(ipi/2)j
Is it possible to find ln(i+j)? Cand I find ln(i+j+ε), where ε^2=0, with ε≠0?
Thank you.
Split-complex numbers can be represented as 2x2 matrices:
a+bj= (a b b a).
So,
j^j= (0 1 ^ (0 1) 1 0) 1 0) = (0 1 1 0) =j
Verify it with Wolfram Alpha.
It seems you possibly made a mistake with sign.
The combination of complex and split-complex numbers is called tessarines, they can be represented as 2x2 matrices
a+bj= (a b b a)
with complex elements a and b.
So, in tessarines, you are correct that i^j=ij. The j^i is
((½)−(e^(−π)/2))j + (e^(−π)/2) + ½