Does this equation have any solutions:


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

  • $\begingroup$ Certainly $z=i$ is one solution. $\endgroup$
    – Shaun Ault
    Commented Jul 10, 2013 at 11:55
  • 3
    $\begingroup$ Please define $i^z$ (or basically $\log(i)$). $\endgroup$ Commented Jul 10, 2013 at 11:55
  • 2
    $\begingroup$ I think general solutions of this equation are represented as Lambert W function $\endgroup$
    – Hanul Jeon
    Commented Jul 10, 2013 at 12:50
  • $\begingroup$ A related problem. $\endgroup$ Commented Jul 11, 2013 at 9:11

2 Answers 2


$$ \frac1z\log\left(\frac1z\right)=\frac1i\log\left(\frac1i\right)=-\frac\pi2 $$ 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, $z=-i$


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