So, what is the principal value of i^{3i}?

[hbweb500]

I am studying complex variables with Brown and Churchill. In it, they define the principal value of $$z^c$$, with both variables complex, to be $$e^{c\; \text{Log }z}$$, where $$\text{Log}$$ is the principle value branch of the complex logarithm.

Now, suppose $$z = i$$ and $$c = 3$$. We know that $$\text{Arg } i = \frac{\pi}{2}$$, so $$z^c = i^3 = e^{3 \pi / 2}$$. But is this really the principal value? Why don't we say $$e^{- \pi/2}$$ is the principal value?

I ask because it seems like that is what the textbook does in one of its examples: it calculates
$$z^c$$ to be something with an angle outside of $$-\pi < \theta \leq \pi$$, and just reduces it without explanation.

So, when finding the principle value of $$z^c$$ after we have done the calculation, or is simply using the principle value branch logarithm enough?

 

[olivermsun]

Log already chooses a principal value by only taking i Arg z for the complex part. Hence it isn't necessary to introduce further conventions to get an principal value for z^c.

Anyhow, your exponential should have an i upstairs I think, so that it doesn't matter whether the exponent is -i pi/2 or i 3 pi/2.

Now if you were to take Arg of the exponential then I suppose you'd have to return a value in (-pi, pi].

 

[HallsofIvy]

Yes, you are missing an "i" in the numerator $i^3= e^{3i\pi/2}= -i$ as you get by straight forward multiplication: $i^3= (i^2)i= -i$.

Doing it as [itex]e^{3 log(i)}[itex], $log(i)= i\pi/2+ 2k\pi i$ so that $i^3= e^{3 log(i)}= e^{3i\pi/2+ 6ki\pi}= e^{3i\pi/2}e^{6ki\pi}$. But e to any <b>even</b> multiple of $i\pi$ is 1 so that <b>all</b> "branches" give the same thing. More generally, any complex number to a <b>positive integer</b> power is single valued.<br /> <br /> But, since you said "$z^c$ with both variables complex", did you mean $i^{3i}$. In that case, $i^{3i}= e^{3i log(i)}$ and now $i= e^{i\pi/2}$ so that $log(i)= i\pi/2+ 2k\pi$ as before, $3i \log(i)= -3\pi/2+ 5k\pi$ and, finally, $i^{3i}= e^{-3\pi}e^{5k\pi}$.<br /> <br /> Taking k= 0 gives $i^{3i}= e^{-3\pi}$ which is the smallest positive value. Normally, the "principal value" of a calculation is the non-real value with smallest argument. When all values are real, it is the smallest positive value.[/itex][/itex]