# Fantastic Imaginary unit!

From Euler's identity : $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ For $$\theta=\frac{1}{2}\pi$$ $e^{i\frac{1}{2}\pi}=i$ Let $$a=e^{\frac{1}{2}\pi}$$ $\Rightarrow a^i=i \Rightarrow a^{a^{a^{a^{...}}}}=i$ Now a question which I'm unable to solve and it is very interesting one! (see tetration if you aren't understanding ${^n i}$) $\lim_{n \to \infty}{^n i}=?$

Note by Zakir Husain
1 year ago

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I have plotted the graph of ${^ni}$ on the complex plain using python: The $\lim_{n \to \infty}{^ni}$ is converging in the darker middle region.

${^{960}i}\approx 0.43828...+0.36059...i$

- 1 year ago

Let the limit be $L$. Does $i^L=L?$

- 1 year ago

Yes

- 12 months ago

BTW, I don’t really understand $n^i,$ although I do know tetration. Could you explain it to me? Thanks :)

- 1 year ago

${^ni}=i^{i^{i^{i^{.^{.^{.}}}}}}$ where there are $n$ $i's$

- 1 year ago

Well that I know... :) but $i^i$ I do not understand :D $i$ times itself $i$ times?

- 1 year ago

See how we evaluate $i^i$:

From Euler's identity (It comes every place you talk about $i$): $e^{i\theta}=\cos(\theta)+i\sin(\theta)$ At $\theta = \frac{1}{2}\pi$ $e^{i\frac{1}{2}\pi}=i$ Raising both sides to $i$ $(e^{i\frac{1}{2}\pi})^i=i^i$ $i^i=(e^{i\frac{1}{2}\pi})^i=e^{\red{i}\frac{1}{2}\pi \red{i}}=e^{-\frac{1}{2}\pi}$

- 1 year ago

Oic. Thank you :)

- 1 year ago

- 1 year ago

Sorry but I can't be any help here since I don't know Calculus that well (I have just started the Fundamentals Course)

- 1 year ago

I don't know tetration , but it seems you must calculate exponents from top to botom . Tower of "a" you equal to "i" is right if at the end of tower is and "i" . To calculate the tower you must consider where braquets are . See tetration explanation.

- 12 months ago

Tetration of a complex number to a infinite height can be done using the Lambert W function. The formula is W(-log(z))/(-log(z)). Here log(i)=iπ/2 and the formula gives the value that Zakir Husain has calculated.

- 12 months ago