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 The Powers that Be (Posted on 2003-11-25)
Is it possible for two complex numbers to have a real exponentiation?

In other words, if x and y are complex (each have the form a+bi), show that x^y can have a real value, or prove that it is impossible.

Note: i is the imaginary value defined as the number that yields -1 when squared. a and b are any real numbers, but b is not 0.

 See The Solution Submitted by DJ Rating: 4.4444 (9 votes)

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 solution | Comment 4 of 19 |
We seek x^y=r where x and y are complex and r is real.

This corresponds to x=r^(1/y), where a real is raised to a complex power to get the x value that we need. I don't have a proof that a real to a complex power would be complex, but it certainly seems plausible.

Trying a few numbers in a calculator that handles complex numbers (I used UBASIC in immediate mode as such a calculator), we can find examples such as:

r=10
OK
x=r^(1/(1+#i))
OK
?x
1.2880188077230996985-2.8880802535510616817#i
OK
?x^(1+#i)
9.9999999999999999917-0.000000000000000001#i
OK

Here the symbol #i represents i. The final result shows a tiny imaginary part due to internal rounding errors.

Analytically:
Let x be 1+i

Let y be log(√2) - (π/4)i

To raise a number to a power using logarithms, use log(x^y)=y log(x). I will be using natural logs here and use the mathematical notation log, rather than engineering notation ln.

(1+i) = √2 e^iπ/4
so
log(1+i)= log(√2) + (π/4)i
(which is why we chose its complex conjugate for y)

so y log(x) = log(x^y) is
log(x^y) = (log(√2) + (π/4)i) (log(√2) - (π/4)i)
= (log(√2))^2 - (iπ/4)^2 = (log(√2))^2 + (π^2)/16

This number is a real number. Raising e to this power (taking its natural antilogarithm) will give a real number which is x^y.

 Posted by Charlie on 2003-11-25 13:56:11

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