Compute i^i: Simplifying Complex Math

[buffordboy23]

Homework Statement

Compute $$i^{i}$$

Homework Equations

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

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.

 

[dirk_mec1]

Try writing i in an exponential form.

 

[nicksauce]

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

 

[arildno]

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

 

[buffordboy23]

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

 

[arildno]

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

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

 

[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.