This is the result of my computation:
\( z=\left(\frac{i}{n}\right)^{\frac{n}{i}}=\left(\frac{i}{n}\right)^{-ni}=\frac{e^{i\frac{\pi}{2}(-ni)}}{n^{-ni}}=e^{\frac{\pi}{2}n} n^{ni}=e^{\frac{\pi}{2}n} e^{n\ln(n)i}=e^{\frac{\pi}{2}n+n\ln(n)i} \)
\( y:=-\ln(z)=-\frac{\pi}{2}n-n\ln(n)i \)
If we now set \( x:=\ln(n)-\frac{\pi}{2}i \) then
\( xe^x=(\ln(n)-\frac{\pi}{2}i)e^{\ln(n)-\frac{\pi}{2}i}=(\ln(n)-\frac{\pi}{2}i)n(-i)=-\ln(n)ni-\frac{\pi}{2}n=y \)
And thatswhy \( W(y)=x \) and \( h(z)=\frac{W(y)}{y}=\frac{x}{y} \).
We see above that \( y=xn(-i) \) and hence \( h(z)=\frac{i}{n} \).
So except some minor differences (and not considering branches) this confirms your conjecture.
Well done Ivars!
\( z=\left(\frac{i}{n}\right)^{\frac{n}{i}}=\left(\frac{i}{n}\right)^{-ni}=\frac{e^{i\frac{\pi}{2}(-ni)}}{n^{-ni}}=e^{\frac{\pi}{2}n} n^{ni}=e^{\frac{\pi}{2}n} e^{n\ln(n)i}=e^{\frac{\pi}{2}n+n\ln(n)i} \)
\( y:=-\ln(z)=-\frac{\pi}{2}n-n\ln(n)i \)
If we now set \( x:=\ln(n)-\frac{\pi}{2}i \) then
\( xe^x=(\ln(n)-\frac{\pi}{2}i)e^{\ln(n)-\frac{\pi}{2}i}=(\ln(n)-\frac{\pi}{2}i)n(-i)=-\ln(n)ni-\frac{\pi}{2}n=y \)
And thatswhy \( W(y)=x \) and \( h(z)=\frac{W(y)}{y}=\frac{x}{y} \).
We see above that \( y=xn(-i) \) and hence \( h(z)=\frac{i}{n} \).
So except some minor differences (and not considering branches) this confirms your conjecture.
Well done Ivars!
