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!
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Certainly $z=i$ is one solution. – Shaun Ault Jul 10 '13 at 11:55
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3Please define $i^z$ (or basically $\log(i)$). – Najib Idrissi Jul 10 '13 at 11:55
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2I think general solutions of this equation are represented as Lambert W function – Hanul Jeon Jul 10 '13 at 12:50
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A related problem. – Mhenni Benghorbal Jul 11 '13 at 09:11
2 Answers
15
$$
\frac1z\log\left(\frac1z\right)=\frac1i\log\left(\frac1i\right)=-\frac\pi2
$$
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 $\log(z)=1.60429091344801115852-7.64719227612459292313\,i$.
robjohn
- 345,667
9
$$i\times \ln(z)=z\times \ln(i)$$ Because $\ln(i)=\frac{1}{2}i\pi$ you have: $\ln(z)=-\frac{1}{2}z\pi$ so, $z=-i$
Mikasa
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Riccardo.Alestra
- 10,546