# I to the power of i

tomas
I am starting imaginary numbers in school and I wondered, what is x=ii

RageSk8
e-.5[pi]
Do you know taylor series? If not, learn them, you won't understand the answer until you do.

-quick edit

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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.

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.

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.

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.

Fair enough. That makes sense.