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?

#Algebra

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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 $e^{i\pi/2}$

Swagat Panda - 3 years, 11 months ago

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

Steven Jim - 3 years, 11 months ago

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

Swagat Panda - 3 years, 11 months ago

@Swagat Panda – Eh, it's all okay. Don't blame yourself.

Steven Jim - 3 years, 11 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.

Antoine G - 2 years, 8 months ago

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

Elethelectric Penguin - 3 years, 10 months ago

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

Steven Jim - 3 years, 10 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 $-1$ .

Md Mehedi Hasan - 3 years, 6 months ago

No it can't be

Biswajit Barik - 3 years, 11 months ago

Can you please give reasons for your opinion?

Steven Jim - 3 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$