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Calculus of j^j, i^j, and j^i (where i^2=−1, j^2=1, j≠−1 and j≠1), and others like Ln(i+j)

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.

Answer by Anixx

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