# Thread: Can the ratio of a non-existent number to an imaginary number....

1. ## Can the ratio of a non-existent number to an imaginary number....

ah, thanks guys

2. Originally Posted by sfspitfire23
e^[(pi)(i)] + 1 = 0
e^[(pi)(i)] = -1
(pi)(i) = ln(-1)
pi = [ln(-1)]/i
pi = C/d where C = a circle's circumference and d = a circle's diameter

Thus:
C/d=[ln(-1)]/i
If:
“C” and “d” must be positive real numbers when finding a circle’s area
ln(-1) = Does not exist
i = √(-1) = Imaginary number
Then:
Can the ratio of a non-existent number to an imaginary number equal the ratio of two real numbers? If not, then can pi still be considered a real transcendental number or is it non-existent and/or imaginary?
Actually, ln(-1) does exist, just not as a real number. You have calculated it correctly. (Actually ln(-1) = i(2n + 1)*pi where n is any integer, but we won't quibble about that.)

-Dan

3. many natural logs of -1 exist. the exponential function isn't 1-1 on the complex plane, so we have multiple choices of domains for a (restricted) inverse.

if we choose the principal branch, then in fact, $\displaystyle \ln(-1) = i\pi$.

see here: Complex logarithm - Wikipedia, the free encyclopedia