Vauled-ness of a complex number to an irrational power

[willybirkin]

Homework Statement

For z complex:

a.) is z^\sqrt{2} a multi-valued function, if so how many values does it have?

b.) Claim: z^\sqrt{2}=e^{\sqrt{2}ln(z)}=e^\sqrt{2}e^ln(z)=ze^\sqrt{2}
Since \sqrt{2} has 2 values, z^\sqrt{2} 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 \sqrt{2}=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

 

[SammyS]

For part b:
What two values does \sqrt{2} have ?

 

[willybirkin]

The positive and negative square roots of 2. Or, if you consider 2 to be complex with imaginary part 0 and real part 2, root(2)e^ik\pi with k=0,1.

 

[Hurkyl]

For the most part,
e^{ab} \neq e^a e^b


Also, whether or not \sqrt{2} is a multi-valued number depends on what precisely you mean by the notation -- two different conventions both make sense. Does your textbook state its convention anywhere?

 

[willybirkin]

Wow, I can't believe I missed that. Well then, after some manipulation that equation ought to become r^\sqrt{2}e^{i\sqrt{2}(\theta+2k\pi)} which unless I'm mistaken shouldn't ever be able to return to the original \theta, making it infinitely valued.