Imaginary isn't real, right?

It might be obvious that $$2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { ... } } } } } }$$ equals 4. So what about $$i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { ... } } } } } }$$? The answer might be $$-1$$, but I'm not sure as $$i$$ is not a real number. Can anyone help?

Note by Steven Jim
3 years, 6 months ago

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I don't know if it's absolutely correct, but I am posting it.

If we write $i$ as $e^{i\pi/2}$, then the given series becomes:

\begin{aligned} & e^{i\pi/2} \sqrt{e^{i\pi/2}\sqrt{e^{i\pi/2}\sqrt{e^{i\pi/2}\sqrt{e^{i\pi/2}} \cdots}}} \\ &= e^{i\pi \left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8} \cdots \right)} \\ & =e^{i\pi \left( \frac{1/2}{1-1/2} \right)} \\ & =\boxed{e^{i\pi}=-1} \end{aligned}

Edit: Sorry for the initial error, I wrote $i=e^{i\pi}$, which was absolutely incorrect. It has been corrected now to $e^{i\pi/2}$

- 3 years, 6 months ago

Thanks for fixing. I've been thinking hard ;)

- 3 years, 6 months ago

Sorry for the inconvenience caused due to it. It was an absolute brainfade.

- 3 years, 6 months ago

Eh, it's all okay. Don't blame yourself.

- 3 years, 6 months ago

I would not be as cavalier: when taking square roots with complex numbers, there is no unique of doing it (and you cannot, as in the reals, say I take the positive one). This is best seen if you try to write $1 = \sqrt{1} = \sqrt{ e^{2i \pi} } = e^ {2 i \pi /2} = e^{i \pi } =-1$ I could also write $e^{i\pi/2} \sqrt{ e^{-3i\pi/2} \sqrt{ e^{i\pi/2} \sqrt{\cdots}}} = e^{ i \pi ( \tfrac{1}{2} - \tfrac{3}{4} + \tfrac{1}{8} - \tfrac{3}{16} \ldots) } = e^{i \pi ( \tfrac{2}{3} - 1)} = e^{i \pi /3}$ and get a totally different answer. In fact by choosing a different square root I every step, I could get pretty much any answer.

- 2 years, 3 months ago

Consider that sqrt(-1) = i not sqrt(i) = -1

- 3 years, 5 months ago

I don't really understand what you wanted to mention. Can you please explain clearer?

- 3 years, 5 months ago

Let $x=i\sqrt{i\sqrt{i\sqrt{i\sqrt{i\sqrt{i\sqrt{i\sqrt{......}}}}}}}\\\Rightarrow x=i\sqrt{x}\\\Rightarrow x^2=i^2\cdot x\\\Rightarrow x^2=-x\\\therefore x=i\sqrt{i\sqrt{i\sqrt{i\sqrt{i\sqrt{i\sqrt{i\sqrt{......}}}}}}}=\boxed{-1}$

I am little confused about the 3rd line where I take square in both side. But I think it can be $-1$.

- 3 years, 1 month ago

No it can't be

- 3 years, 6 months ago