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If i can find the the value of the expression in lhs. Then i can find the correct option. But i am unable to find the value of expression on lhs which is i^i^i^i^... Upto infinity . How to find that. I tried to convert this to equation i^m=m but i cant proceed further.

marked as duplicate by Gerry Myerson, Did, user302797, Shailesh, amWhy Feb 21 at 0:24

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up vote 3 down vote accepted

Every level of exponentiation is multi-valued. Suppose we always pick the branch $i=e^{i\pi/2}$ in every level. Then we have

$$(e^{i\pi/2})^{\alpha+i\beta}=e^{-\pi\beta/2}\left(\cos\frac{\pi\alpha}{2}+i\sin\frac{\pi\alpha}{2}\right)=\alpha+i\beta.$$

Two simultaneous equations for real $\,\alpha\,$ and $\,\beta\,$ are established by comparing the real and imaginary parts. We have

$$\left\{\begin{array}{l} e^{-\pi\beta/2}\cos\displaystyle\frac{\pi\alpha}{2}=\alpha,\\ \ \\ e^{-\pi\beta/2}\sin\displaystyle\frac{\pi\alpha}{2}=\beta.\\ \end{array}\right.$$

Therefore $\,\alpha^2+\beta^2=e^{-\pi\beta}\,$ and $\,\frac{\beta}{\alpha}=\tan\frac{\pi\alpha}{2}$ can be found without numerically solving for $\,\alpha\,$ and $\,\beta$. Their numerical values are given by $\,\alpha=0.4383\,$ and $\,\beta=0.3606$.

Let $i^{i^{i^{\cdots}}}=\alpha+i\beta$ so $i^{\alpha+i\beta}=\alpha+i\beta$, now take logs of this \begin{eqnarray*} (\alpha+i\beta)i \frac{\pi}{2} =\ln(\alpha^2+\beta^2)+i \tan^{-1}( \beta / \alpha). \end{eqnarray*} Consider the imaginary parts ....

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