Negative ^ Irrational

What is the value of (1)2(-1)^{\sqrt{2}}

Note by Chris Sapiano
4 months ago

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Let (1)2=A(-1)^{\sqrt{2}}= A

Then on taking the natural logarithm on both sides:

2ln(1)=ln(A)\sqrt{2}\ln(-1) = \ln(A) Which implies:

2ln(i2)=2ln(eiπ)=2iπ=ln(A)\sqrt{2}\ln(i^2) = \sqrt{2}\ln(e^{i\pi})=\sqrt{2}i\pi= \ln(A)

Or, A=eiπ2=cos(π2)+isin(π2)A = e^{i\pi\sqrt{2}} = \cos(\pi\sqrt{2}) + i \sin(\pi\sqrt{2})

So, the result is a complex number.

Note: i=1i = \sqrt{-1} eiθ=cos(θ)+isin(θ)e^{i\theta}= \cos(\theta) + i\sin(\theta)

Karan Chatrath - 4 months ago

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Not quite right. i2i^2 can be equal to e3iπ e^{3i \pi} as well. So the answer could also be cos(3π2)+isin(3π2) \cos(3\pi \sqrt2) + i \sin(3\pi \sqrt2) .

Pi Han Goh - 4 months ago

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Good observation. The answer can be generalised for odd multiples of π\pi. However, my intent was to only show that the result of raising a negative number to an irrational exponent is complex. Generalisation was not the intent of the comment.

The uniqueness of the resulting complex number is something to ponder on. Is there a better way of approaching such a computation?

Karan Chatrath - 4 months ago

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@Karan Chatrath I really don't know if there's a better way.

Pi Han Goh - 4 months ago

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@Karan Chatrath If you were to generalize it, how about e^ki(pi)=.... for k is an odd integer. Forgive me, I do not know how to use the math symbols on this yet.

Dacota Sprague - 4 months ago

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@Dacota Sprague Yes, you are correct, this is how I would generalise it.

Karan Chatrath - 4 months ago

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@Karan Chatrath I think this answer is the principal one.

Ruilin Wang - 3 months, 3 weeks ago

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@Ruilin Wang Hello! Could you explain why?

Karan Chatrath - 3 months, 3 weeks ago

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@Karan Chatrath Oh. I just meany that because this is the one without any constants, so it should be the principal one, like the principal cube root of unity is 1.

Ruilin Wang - 3 months, 3 weeks ago

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