I^i = ?

1. Aug 29, 2008

buffordboy23

1. The problem statement, all variables and given/known data

Compute $$i^{i}$$

2. Relevant equations

$$z = x + iy$$
$$z = r \left( cos\theta + isin\theta \right)$$

3. The attempt at a solution

I tried to manipulate the equations for substitution to no avail. I think that the answer is 1, since this is what my calculator told me, but I haven't got any more ideas to proceed with the substitutions for i. Experimenting with different ideas on a calculator, I noticed that $$n^{i}$$ , where $$n$$ is a positive integer, gave a complex number of the form, $$a = b + ic$$, where $$a^{2} = 1$$ always, and where $$b$$ decreases and $$c$$ increases as the integer $$n$$ increases.

2. Aug 29, 2008

dirk_mec1

Try writing i in an exponential form.

3. Aug 29, 2008

nicksauce

i^i = e^{i*log(i)}

4. Aug 29, 2008

arildno

Utilize the exponential representation(s) of the imaginary unit,
$$i=e^{i(\frac{\pi}{2}+2n\pi}$$
where n can be any integer.

5. Aug 29, 2008

buffordboy23

I solved the problem...thanks for the help.

6. Aug 30, 2008

arildno

Really?
Could you show us the answer you arrived at?

7. Aug 30, 2008

buffordboy23

Sure...

From Euler's formula,

$$e^{i\theta} =\left( cos\theta + isin\theta \right)$$

Let $$\theta = \frac{\pi}{2}+2n\pi$$, where n is an integer. Therefore,

$$e^{i\left(\frac{\pi}{2}+2n\pi\right)} = i$$

for any integer n. Taking both sides to ith power and noting that $$i^{2} = -1$$ gives

$$i^{i} = e^{-\left(\frac{\pi}{2}+2n\pi\right)}$$

Very interesting result....since $$i^{i}$$ can be comprised of an infinite number of elements. Again, thanks for the help.