# Does $z^i=i^z$ have any solutions, beside $z=i$?

Does this equation have any solutions:

$$\large{z^i=i^z}$$

Putting polar form of $z$ is better for LHS, But rectangular form is suitable for RHS ! What to do? Thanks!

• Certainly $z=i$ is one solution. Commented Jul 10, 2013 at 11:55
• Please define $i^z$ (or basically $\log(i)$). Commented Jul 10, 2013 at 11:55
• I think general solutions of this equation are represented as Lambert W function Commented Jul 10, 2013 at 12:50
• Commented Jul 11, 2013 at 9:11

$$\frac1z\log\left(\frac1z\right)=\frac1i\log\left(\frac1i\right)=-\frac\pi2$$ Thus, there are many solutions, one for each branch of the Lambert W function: $$z=e^{-\mathrm{W}\left(-\pi/2\right)}$$ One solution is Exp[-LambertW[0,-Pi/2]] which is -i.

Another is Exp[-LambertW[-1,-Pi/2]] which is i.

However, another is N[Exp[-LambertW[1,-Pi/2]],20] which is

1.0213233161306520062 - 4.8683538060775645979 i

Of course, the value of $z^i$ depends on which branch of $\log(z)$ is used. In the computation above, I have taken $\log(i)=\pi i/2$, so we have a clear definition of $$i^z=e^{\pi iz/2}$$ However, $\log(z)$ is determined only mod $2\pi i$ and that affects $z^i$ by a factor of $e^{2\pi k}$ for $k\in\mathbb{Z}$.

For example, for $z=1.0213233161306520062-4.8683538060775645979\,i$, we use $\log(z)=1.60429091344801115852-7.64719227612459292313\,i$.

$$i\times \ln(z)=z\times \ln(i)$$ Because $\ln(i)=\frac{1}{2}i\pi$ you have: $\ln(z)=-\frac{1}{2}z\pi$ so, $z=-i$