What is the significance of x=ii in imaginary numbers?

  • Thread starter tomas
  • Start date
  • Tags
    Power
In summary, x = i^i can be solved using complex logarithms and the exponentiation rule to obtain x = exp(-Pi/2 - 2*Pi*k), where k is an integer. This yields an infinite number of solutions and it is important to specify whether the principal value or all possible solutions are desired.
  • #1
tomas
I am starting imaginary numbers in school and I wondered, what is x=ii
 
Mathematics news on Phys.org
  • #2
e-.5[pi]
Do you know taylor series? If not, learn them, you won't understand the answer until you do.

-quick edit
 
Last edited by a moderator:
  • #3
x = exp(-Pi/2)

x=ii
Taking logs to base e

ln(x) = i*ln(i)
Since ln(i)= i*Pi/2, ln(x)=-Pi/2, so raising both sides to the power of e we obtain.
x = exp(-Pi/2)

Explaination of ln(i) = -i*Pi/2

if w is a complex number such that exp(w) = z, then w = ln|z|+i*arg(z). Since |i|=1, and arg(i)=Pi/2, ln(i) turns out to be -Pi/2.
 
  • #4
When working with complex exponents, remember that you will get an infinite number of answers. The standard way of solving log problems doesn't really apply here. This is how you would solve this problem:

x=i^i => x = exp(i*log(i))
This is a definition from Complex Analysis
log(i) = Log|i| + i*arg(i)
This is the definition of log(z) where z is complex. Note it is not quite the same as ln(x).
Log(x) is a real function and works exactly the same as ln(x), its just a different terminology used in math.
arg(i) is definited as the angle the 'imaginary vector' makes with the positive real axis.
|i| is the magnitude of i (absolute value), which is 1.
So, log(i) = Log(1) + i*arg(i) = 0 + i*Pi/2 + i*2*Pi*k where k is an integer (by definition of arg(z))
Thus,x = i^i = exp(i*i(Pi/2 + 2*Pi*k) = exp(-Pi/2 - 2*Pi*k) k element of Z/ (The Integers)
Which provides for an infinite number of solutions.

Hope that helps.
 
  • #5
Why can't we use the principal logarithm? Apostol uses it in the analysis textbook I use, and I'm more than happy to follow his example.
 
  • #6
It is certainly reasonable to use the principle value of the log, if that is what is asked for. However, if someone wants x=i^i, without specifying they only want the principle value (k=0), you need to specify all possible solutions. It would be like solving a standard quadratic, and only providing the positive solution. Its still correct, but incomplete.
 
  • #7
Fair enough. That makes sense.
 

What is "I to the power of i"?

"I to the power of i" is a mathematical expression that involves raising the imaginary number "i" to the power of itself. "i" is the square root of -1 and is often used in complex numbers and in solving certain equations.

What is the value of "I to the power of i"?

The value of "I to the power of i" is not a real number, but a complex number with a real and imaginary component. It is approximately equal to 0.2079 + 0.8626i.

What is the significance of "I to the power of i"?

"I to the power of i" has many important applications in mathematics, physics, and engineering. It is used in solving differential equations, in signal processing, and in the study of quantum mechanics.

Can "I to the power of i" be simplified?

Yes, "I to the power of i" can be simplified using Euler's formula, which states that e^(ix) = cos(x) + i*sin(x). By substituting i for x in this formula, we can simplify "I to the power of i" to e^(-π/2).

What is the relationship between "I to the power of i" and the complex unit circle?

When graphed on the complex plane, "I to the power of i" lies on the unit circle, which is a circle with a radius of 1 centered at the origin. This is because its value can be represented as e^(-iπ/2), which is the point on the unit circle at 90 degrees or π/2 radians.

Similar threads

Replies
2
Views
1K
Replies
8
Views
1K
Replies
10
Views
1K
  • General Math
Replies
3
Views
724
Replies
5
Views
716
  • General Math
Replies
29
Views
3K
  • General Math
Replies
11
Views
2K
Replies
2
Views
1K
  • General Math
Replies
1
Views
1K
Replies
12
Views
2K
Back
Top