Let's say we want to evaluate

\[ \frac{ (-1) ^ {-i} } { i ^ i } = ( a \sqrt{a} ) ^ \pi. \]

What are wrong with the following methods?

3rd Method:

Raise both sides to the power of 4

\( \large \frac { ((-1)^4)^{-i}}{(i^4)^i} = a^{6\pi} \), then \( \frac {1^{-i}}{1^i} = 1 = a^{6\pi} \)

With that, we have \(a = 1 \)

4th method:

Raise both sides to the power of 3

\( \large \frac { ((-1)^3)^{-i}}{(i^3)^{i}} = \frac {((-1)^{-1})^{i}} {(-i)^i } = \frac {(-1)^i}{(-i)^i} = \left ( \frac 1i \right )^i = (-i)^i = a^{9\pi /2} \)

\( (-i)^{-i} = a^{-9\pi /2} \) or \( e^{-\pi /2} = a^{-9\pi /2} \), then \(a = e^{1/9} \).

5th Method:

Raise both sides to the power of 5

\( \large \frac { ((-1)^5)^{-i}}{(i^5)^i} = a^{15\pi /2} \)

\( \large \frac {(-1)^{-i}}{i^i} = \left ( \frac {-1}{i} \right )^i = i^i = e^{-\pi /2} = a^{15\pi /2} \)

\(a = e^{-1/15} \).

6th Method:

Take square root to both sides of the equation, with numerator of fraction becomes \( (-1)^{-i/2} = ( \sqrt{-1} )^{-i} = i^{-i} \)

\( \large \frac {i^{-i} }{i^{i/2}} = a^{3\pi /4} \)

\( i^{-3\pi /2} = a^{3\pi /4} \)

\( i^{-1} = a^{1/2} \)

\( -i = a^{1/2} \)

\( a = (-i)^2 = -1 \).

7th Method:

Knowing that \(i^2 = -1 \), we could have \( (-1)^{-i} = i^{-i/2} \)

\( \large \frac { i^{-i/2} }{ i^i} = i^{-3i/2} = a^{3\pi /2} \)

\( i^{-i} = a^\pi, i^i = a^{-\pi}, e^{-\pi /2} = a^{-\pi} \)

\( a = e^2 \).

8th Method:

Knowing that \(i^2 = -1 \), we could have \( i^i = (-1)^{i/2} \)

\( \large \frac {(-1)^{-i} }{(-1)^{i/2}} = (-1)^{-3i/2} = a^{3\pi /2} \)

\( (-1)^{i} = a^{\pi} \)

\( (e^{i \pi} )^{i} = a^\pi \)

\( a = e^{i^2} = e^{-1} \).

9th Method:

Like the 8th Method

\( (-1)^{i} = a^{\pi} \)

\( i^{i/2} = a^{\pi} \)

\( i^i = a^{2 \pi} \)

\( e^{-\pi /2} = a^{2 \pi} \)

\( a = e^{-1/4} \).

Original problem. See the first 2 methods there.

## Comments

Sort by:

TopNewestAll of the examples involve complex exponentiation of negative numbers, or, in the general case, complex exponentiation of complex numbers, which can have an infinity of values instead of an unique one. Consequently, we can't rely on ordinary identities and algebraic properties of exponentiation, as for example

\({ \left( { a }^{ b } \right) }^{ c }={ { \left( { a }^{ c } \right) } }^{ b }\)

is not true if \(a=-1\), \(b=3\), \(c=i\).

Is there a "correct" answer to this? Outside of convention, no. This infinitude of values is an inherent property of complex exponentiation of complex numbers. So, special care has to be taken to avoid running into a multitude of "fallacies" as given above. – Michael Mendrin · 2 years, 2 months ago

Log in to reply

\((e^{i\pi})^i=a^{-\pi}\)

\(\Rightarrow a=e\)

But:

\((e^{i\pi})^i=(e^{3i\pi})^i=a^{-\pi}\)

\(\Rightarrow a=e^{3}\)

\(\Rightarrow e=e^{3}\)

– Raghav Vaidyanathan · 2 years, 2 months agoLog in to reply

In method 6,you have made calculation mistake in step 3 – Akhil Bansal · 2 years ago

Log in to reply

Sir, in the method 3, you can't just raise any power to 2m. for eg. x^2=(-x)^2. You will be including roots of both positive and negative functions. I hope I am right. If wrong please do correct me. – Aditya Kumar · 2 years, 2 months ago

Log in to reply