real number value for imaginary unit

**rainer** · September 26th 2009, 11:09 AM

Since $\cos(i)$ is a real number (1.54308...)

Does this mean that I can substitute "i" as follows?

$i=\arccos(1.54308...)$

If not, why not?

Also, since $i^i=\frac{1}{e^\frac{\pi}{2}}$ does this mean that

$i=\arccos(1.54308...)=\left(\frac{1}{e^\frac{\pi}{ 2}}\right)^\frac{1}{i}$

?

**Dark Sun** · September 26th 2009, 01:29 PM

The first one follows. If $\mbox{cos }(i)=1.54308...$ , then $i=\mbox{arccos}(1.54308...)$

The second equation follows as well. Provided that $i^i=\frac{1}{e^{\pi /2}}$ , then by taking the $\frac{1}{i}$ power of both sides of the equation, we have:

$(\frac{1}{e^{\pi /2}})^{\frac{1}{i}}=i=\mbox{arccos}(1.54308...)$ .

I'm not sure what you are going to use these facts to solve, but each of these steps are correct. Are you certain that $\mbox{cos }(i)=1.54308...$ ? Are we speaking in terms of hypothetical? Imaginary numbers are tricky, it would help if I knew where you were coming from.

I suspect that each of these steps are correct, but that the premise may be in error.

**rainer** · September 28th 2009, 07:36 AM

According to wikipedia's page on the imaginary unit:

$\cos(i) = \cosh(1)= (1.54308...)$

The page doesn't show how this is derived.

I don't have a particular use in mind for this info. I was just cruising wikipedia and this jumped out at me.

Even if cos(i) is a real number, I guess this does not give us a real number value for "i" itself. Acos(cos(i)) does not exist, or is imaginary.

Same with $i^i$ .

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