Simplified imaginary unit to the power of imaginary unit

[Leo Authersh]

Where is my simplification wrong?

 

[BvU]

Do I distinguish a $(-1)^{1/2} = -{1\over 2}$ in the first step ?

 

[Leo Authersh]

Do I distinguish a $(-1)^{1/2} = -{1\over 2}$ in the first step ?

I wrote it based on the general formula for real numbers that $((a)^b)^c = a^{bc}$

 

[mfb]

That formula is (a) problematic for complex numbers and (b) doesn't lead to -1/2 even if you apply it.

 

[Leo Authersh]

That formula is (a) problematic for complex numbers and (b) doesn't lead to -1/2 even if you apply it.

$(((64)^{1/2})^{-4})^ {1/2} = (64^{1/2})^{-4/2} = 1/64$
I wrote it on this basis.

 

[jbriggs444]

When you replaced $i^i$ with ${((-1)^{(1/2)})}^{((-1)^{(1/2)})}$, you forgot the parentheses around the "i" in the exponent.

Exponentiation is not associative. $(a^b)^c$ is not, in general, equal to $a^{(b^c)}$.

 

[Leo Authersh]

When you replaced $i^i$ with ${((-1)^{(1/2)})}^{((-1)^{(1/2)})}$, you forgot the parentheses around the "i" in the exponent.

Exponentiation is not associative. $(a^b)^c$ is not, in general, equal to $a^{(b^c)}$.

Hi,

I think this will make it clearer.

 

[jbriggs444]

All of which is irrelevant because your calculation does not have that form.

 

[Leo Authersh]

When you replaced $i^i$ with ${((-1)^{(1/2)})}^{((-1)^{(1/2)})}$, you forgot the parentheses around the "i" in the exponent.

Exponentiation is not associative.

All of which is irrelevant because your calculation does not have that form.

And that is my question. Why the general formula becomes irrelevant when 4096 is replaced with -1.

 

[jbriggs444]

And that is my question. Why the general formula becomes irrelevant when 4096 is replaced with -1.

You misunderstand. It's not the -1 that's the problem. It's the parentheses.

Edit:
${(a^b)}^c$ is equal to $a^{(bc)}$

But obviously,

$a^{(b^c)}$ is not equal to $a^{(bc)}$ [unless $b^c$ just happens to equal $b \times c$ or a just happens to be equal to 1]

 

[Leo Authersh]

You misunderstand. It's not the -1 that's the problem. It's the parentheses.

Edit:
${(a^b)}^c$ is equal to $a^{(bc)}$

But obviously,

$a^{(b^c)}$ is not equal to $a^{(bc)}$ [unless $b^c$ just happens to equal $b \times c$ or a just happens to be equal to 1]

Thank you. Now I understood why exponentials are not left associative. But since apparently they are all right associative, I'm confused because my simplification gives me the deception of being right associative.

 

[mfb]

Where do you see something of the type of $(a^b)^c$ in $(-1)^{1/2}$? What would a,b, and c be?

 

[jbriggs444]

Thank you. Now I understood why exponentials are not left associative. But since apparently they are all right associative, I'm confused because my simplification gives me the deception of being right associative.

Personally, I can never remember whether a tower of exponents is supposed to be evaluated from left to right or from right to left. I'm pretty sure that I've seen it both ways over the years. But it does not matter. If you are the one writing down the formula, you can use parentheses to be absolutely sure that it evaluates the way you want it to.

So let's step through it slowly.

Start with $$i^i$$
Replace each i with $({(-1)}^{(1/2)})$. The parentheses are there to ensure proper evaluation order. $${({(-1)}^{(1/2)})}^{({(-1)}^{(1/2)})}$$
Now look at that expression and try to find something with the form of ${(a^b)}^c$. You originally thought you had something with:
a=${({(-1)}^{(1/2)})}$
b=$(-1)$
c=$(1/2)$
But the parentheses mess that up. It's not of the right form.

 

[Leo Authersh]

Where do you see something of the type of $(a^b)^c$ in $(-1)^{1/2}$? What would a,b, and c be?

Now, I have understood. Thank you for your answers.

 

[Svein]

Messing around with complex numbers - let's see: i=e^{i\frac{\pi}{2}}, so i^{i}=(e^{i\frac{\pi}{2}})^{i}=e^{i\frac{\pi}{2}\cdot i}=e^{i \cdot i\frac{\pi}{2}}=e^{-\frac{\pi}{2}}.

 

[FactChecker]

Messing around with complex numbers - let's see: i=e^{i\frac{\pi}{2}}, so i^{i}=(e^{i\frac{\pi}{2}})^{i}=e^{i\frac{\pi}{2}\cdot i}=e^{i \cdot i\frac{\pi}{2}}=e^{-\frac{\pi}{2}}.

Conclusion -- When dealing with complex exponents, consider the complex exponential function.

 

[Infrared]

Messing around with complex numbers - let's see: i=e^{i\frac{\pi}{2}}, so i^{i}=(e^{i\frac{\pi}{2}})^{i}=e^{i\frac{\pi}{2}\cdot i}=e^{i \cdot i\frac{\pi}{2}}=e^{-\frac{\pi}{2}}.

It's a little bit tricker than this: $i=e^{\frac{5\pi}{2}i}$ is also true and, following your steps, $i^i=(e^{\frac{5\pi}{2}i})^i=e^{-\frac{5\pi}{2}}$. The issue is that it's hard to define exponentiation for complex numbers. We want to say $z^w=\exp(w\log(z))$ (for $z\neq 0$ of course), but how to define the complex logarithm? For $z=re^{i\theta}$, we can't have $\log(z)=\log(r)+i\theta$ since $\theta$ is only defined up to a multiple of $2\pi$, unless we restrict our allowed $\theta$, say to be in $[0,2\pi)$. But this makes $\log$ discontinuous and causes exponent rules to fail, as seen above.

Alternatively, you can just give up on $\log$ being a function, and instead have it be defined only up to multiples of $2\pi i$. But then our exponential $z^w=\exp(w\log(z))$ becomes defined only up to factors of $\exp(2\pi iw)$. With $w=i$, we get back to the original problem.

This sort of problem leads to the theory of Riemann surfaces.

 

[FactChecker]

It's a little bit tricker than this: $i=e^{\frac{5\pi}{2}i}$ is also true and, following your steps, $i^i=(e^{\frac{5\pi}{2}i})^i=e^{-\frac{5\pi}{2}}$. The issue is that it's hard to define exponentiation for complex numbers. We want to say $z^w=\exp(w\log(z))$ (for $z\neq 0$ of course), but how to define the complex logarithm? For $z=re^{i\theta}$, we can't have $\log(z)=\log(r)+i\theta$ since $\theta$ is only defined up to a multiple of $2\pi$, unless we restrict our allowed $\theta$, say to be in $[0,2\pi)$. But this makes $\log$ discontinuous and causes exponent rules to fail, as seen above.

Alternatively, you can just give up on $\log$ being a function, and instead have it be defined only up to multiples of $2\pi i$. But then our exponential $z^w=\exp(w\log(z))$ becomes defined only up to factors of $\exp(2\pi iw)$. With $w=i$, we get back to the original problem.

This sort of problem leads to the theory of Riemann surfaces.

Good point. It's true that you need to define a branch of the logarithm. The principle branch being the most standard. So there are more than one answer to the question. But the answer given is definitely one possible answer.

 

[Svein]

It's a little bit tricker than this: $i=e^{\frac{5\pi}{2}i}$ is also true and, following your steps, $i^i=(e^{\frac{5\pi}{2}i})^i=e^{-\frac{5\pi}{2}}$. The issue is that it's hard to define exponentiation for complex numbers. We want to say $z^w=\exp(w\log(z))$ (for $z\neq 0$ of course), but how to define the complex logarithm? For $z=re^{i\theta}$, we can't have $\log(z)=\log(r)+i\theta$ since $\theta$ is only defined up to a multiple of $2\pi$, unless we restrict our allowed $\theta$, say to be in $[0,2\pi)$. But this makes $\log$ discontinuous and causes exponent rules to fail, as seen above.

Alternatively, you can just give up on $\log$ being a function, and instead have it be defined only up to multiples of $2\pi i$. But then our exponential $z^w=\exp(w\log(z))$ becomes defined only up to factors of $\exp(2\pi iw)$. With $w=i$, we get back to the original problem.

This sort of problem leads to the theory of Riemann surfaces.

Yes, of course. But I just wanted to show that a purely imaginary exponent of a purely imaginary number has a real value. What you have done is to show that it has several real values, which is true for a rather large class of complex functions.