# Negative ^ Irrational

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

Note by Chris Sapiano
8 months, 3 weeks ago

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

Then on taking the natural logarithm on both sides:

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

$\sqrt{2}\ln(i^2) = \sqrt{2}\ln(e^{i\pi})=\sqrt{2}i\pi= \ln(A)$

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

So, the result is a complex number.

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

- 8 months, 3 weeks ago

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

- 8 months, 2 weeks ago

Good observation. The answer can be generalised for odd multiples of $\pi$. 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?

- 8 months, 2 weeks ago

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

- 8 months, 2 weeks ago

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.

- 8 months, 2 weeks ago

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

- 8 months, 2 weeks ago

Try SearchOnMath.

- 8 months, 2 weeks ago

I think this answer is the principal one.

- 8 months, 1 week ago

Hello! Could you explain why?

- 8 months, 1 week ago

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.

- 8 months, 1 week ago