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

  • Context: Graduate 
  • Thread starter Thread starter hbweb500
  • Start date Start date
  • Tags Tags
    Definition Value
Click For Summary
SUMMARY

The principal value of \(i^{3i}\) is determined using the formula \(e^{c \; \text{Log } z}\), where \(z = i\) and \(c = 3\). The calculation shows that \(i^{3i} = e^{-3\pi}\), which is the smallest positive value. The discussion clarifies that while \(i^3 = -i\) can be derived from straightforward multiplication, the principal value is defined as the non-real value with the smallest argument. The use of the principal value branch of the complex logarithm is essential for accurate calculations.

PREREQUISITES
  • Understanding of complex variables and functions
  • Familiarity with the complex logarithm and its principal value
  • Knowledge of Euler's formula and its application in complex exponentiation
  • Basic skills in manipulating complex numbers and their arguments
NEXT STEPS
  • Study the properties of the complex logarithm in detail
  • Learn about the implications of branch cuts in complex analysis
  • Explore the application of Euler's formula in complex exponentiation
  • Investigate the concept of principal values in complex functions
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on complex analysis, as well as educators teaching complex variables and their applications.

hbweb500
Messages
40
Reaction score
1
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?
 
Physics news on Phys.org
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].
 
Last edited:
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 e^{3 log(i)}, 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.
 
Last edited by a moderator:

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
8
Views
3K