Complex enough for you?

Algebra Level 3

Find the principle real value for the given expression below:

${ (-1) }^{- i }$

\frac{162}{7}

\sqrt{ \pi }

\sqrt{ 2 }

-1

e^{ \pi }

\frac{1}{e}

1

2 solutions

Brian Charlesworth
Mar 9, 2015

Note that $-1 = e^{i\pi}$ , so $(-1)^{-i} = (e^{i\pi})^{-i} = e^{-i^{2}\pi} = \boxed{e^{\pi}}.$

But when I tried it using a calculator to see where I was wrong, the answer was 23.1428571428571429 which is 162/7

Aryan Gaikwad - 6 years, 3 months ago

$e^{\pi}$ is very close to $\frac{162}{7}$ but it's not exactly the same. $e^{\pi}$ is in fact a transcendental number, a special type of irrational number, and hence cannot be represented by a fraction.

Brian Charlesworth - 6 years, 3 months ago

Seth Lovelace
Mar 10, 2015

Well here goes:

${ (-1) }^{ -i }={ e }^{ -ilog(-1) }\\ log(z)=log(r)+i(\theta +2\pi n)\\ log(-1)=log(1)+i(\pi +2\pi n)\\ log(-1)=i\pi \\ { (-1) }^{ -i }={ e }^{ -i(i\pi ) }\\ ({ -1) }^{ -i }={ e }^{ \pi }$

And while 162/7 and ${ e }^{ \pi }$ are quite close numerically, they are definitely not the same quantity. Because we are focusing on the principle real value, the ${2\pi n}$ was dropped. This is just a bit more of a start from scratch approach to show where the value comes from. Brian's approach by starting with Euler's identity is equally as valid.

Note: In this context, log(x) is assumed to be a logarithm of base e, not base 10.

Complex enough for you?

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