# What is i^i?

1. Dec 13, 2004

### Hyperreality

What is i^i?? And how do we find it?

2. Dec 13, 2004

### Hurkyl

Staff Emeritus
By the definitions!

For complex numbers, exponentiation is defined by:

$$z^w := \exp(w \log z)$$

For the principal value of the exponential, you use the principal value of the logarithm.

3. Dec 13, 2004

### dextercioby

Quite remarkable,Hurkyl,that both $i^{i}$ and $i^{\frac{1}{i}}$ are real numbers.

Yap,sometimes mathematics offers surprises...

4. Dec 13, 2004

### Tide

Alternate HINT:

$$i^i = \left(e^{i\pi /2}\right)^i$$

5. Dec 13, 2004

### dextercioby

:rofl: That's not a "HINT",that's the SOLUTION!!Hurkyl gave a hint.Anyway,i hope it helps him... :rofl:

6. Dec 13, 2004

### futb0l

7. Dec 14, 2004

### tongos

well, what stumps me is how to find like (i+5)^(i+5)?????

8. Dec 14, 2004

### Tide

HINT: Follow the hints offered above!

9. Dec 15, 2004

### Tsss

The fact that $$(a^b)^c=a^{bc}$$ is not right with complex numbers.

As a matter of fact, let z a complex number,
$$e^z=e^{\frac{2i\pi z}{2i\pi}}=(e^{2i\pi})^{\frac{z}{2i\pi}}=1^{\frac{z}{2i\pi}}=1$$
there is a problem.

10. Dec 15, 2004

### cronxeh

or, if you are an engineer, just open matlab and do i^i will save you the valuable time

11. Dec 15, 2004

### Tide

That's slick but

$$1^{\frac{z}{2\pi i}}=1$$

only if $1 = e^{0i}$ (on the LHS) but you explicitly took $1 = e^{2\pi i}$ and used a different expression of 1 in your final step. $1^z$ will be 1 only if arg z = 0.

12. Dec 15, 2004

### learningphysics

So when manipulating complex numbers we can't simply make a substitution and say:
$$1^\frac{z}{2\pi i}=(e^{(0i)})^{\frac{z}{2\pi i}}=e^0=1$$ ??

Which step above is illegal when manipulating complex numbers?

Is there a webpage, which states these types of situations...

13. Dec 15, 2004

### Tide

That is a valid root of 1 and there is nothing wrong with what you did. Tsss's problem arose because he inconsistently expressed the same number in two different ways going from one step to the next.

$$\left(e^{0 i}\right)^{1/2} = e^0 = 1$$

and

$$\left(e^{2\pi i}\right)^{1/2} = e^{i\pi} = -1$$

are distinct roots of 1 but if I applied Tsss's method to the second I could find only one square root of 1. Obviously, squaring either 1 or -1 both give 1.