- #1

willybirkin

- 3

- 0

## Homework Statement

For z complex:

a.) is z

^{[itex]\sqrt{2}[/itex]}a multi-valued function, if so how many values does it have?

b.) Claim: z

^{[itex]\sqrt{2}[/itex]}=e

^{[itex]\sqrt{2}[/itex]ln(z)}=e

^{[itex]\sqrt{2}[/itex]}e

^{ln(z)}=ze

^{[itex]\sqrt{2}[/itex]}

Since [itex]\sqrt{2}[/itex] has 2 values, z

^{[itex]\sqrt{2}[/itex]}is 2 valued.

Is this correct? If not, correct it.

## Homework Equations

## The Attempt at a Solution

For part a I would intuitively assume that it is infinitely-valued since z

^{1/n}has n values and [itex]\sqrt{2}[/itex]=1+4/10+1/100+4/1000... so its denominator is approaching infinity. But this isn't exactly mathematically sound reasoning since referring to the "denominator" of an irrational number doesn't make any sense.

For part b nothing really jumps out at me as being incorrect, except that it's conclusion disagrees with my belief for part a. It seems like all the steps are mathematically sound, unless there is some reason that you can't simplify e

^{ln(z)}to z, in which case ln(z) being infinitely-valued would make the whole thing infinitely-valued