Imaginary isn't real, right?

It might be obvious that 222222...2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { 2\sqrt { ... } } } } } } equals 4. So what about iiiiii...i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { i\sqrt { ... } } } } } } ? The answer might be 1-1, but I'm not sure as ii is not a real number. Can anyone help?

Note by Steven Jim
2 years, 3 months ago

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

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

eiπ/2eiπ/2eiπ/2eiπ/2eiπ/2=eiπ(12+14+18)=eiπ(1/211/2)=eiπ=1 \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=eiπi=e^{i\pi} , which was absolutely incorrect. It has been corrected now to eiπ/2e^{i\pi/2}

Swagat Panda - 2 years, 3 months ago

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

Steven Jim - 2 years, 3 months ago

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

Swagat Panda - 2 years, 3 months ago

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

Steven Jim - 2 years, 3 months ago

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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=1=e2iπ=e2iπ/2=eiπ=1 1 = \sqrt{1} = \sqrt{ e^{2i \pi} } = e^ {2 i \pi /2} = e^{i \pi } =-1 I could also write eiπ/2e3iπ/2eiπ/2=eiπ(1234+18316)=eiπ(231)=eiπ/3 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 - 1 year ago

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

Elethelectric Penguin - 2 years, 3 months ago

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

Steven Jim - 2 years, 3 months ago

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Let x=iiiiiii......x=ixx2=i2xx2=xx=iiiiiii......=1x=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 1-1.

Md Mehedi Hasan - 1 year, 10 months ago

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No it can't be

Biswajit Barik - 2 years, 3 months ago

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Can you please give reasons for your opinion?

Steven Jim - 2 years, 3 months ago

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