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.

Your example is great. It shows what happens if one doesn't carefully define what z^c is for complex z and c. So let's take a stab at defining things and see what happens then.

Let x,y,r,t be real, z=x+iy, and let n be an integer. Denote pi by P.

Assume we know e^(it)=cos(t)+i*sin(t), e^z=(e^x)(e^iy), and e^z is never 0. We should perhaps write exp(z) instead of e^z, since it will turn out that this e^z is a specially designated one of the possible values that (2.718...)^z could be defined to have. When we write e^z below we always mean the one we just defined here.

One can easily show that e^(z+z')=(e^z)(e^z')and e^(z+i2nP)=e^z.

If equation e^z=z' has a solution z, then z+i2nP is also a solution for any n. Each such solution is a logarithm of z'. To have a well-defined logarithm function, we need a rule to pick out just one solution. We will do that like this:

Define the function Log by Log(z)=log(r)+it, where z=r*e^(it), log is the ordinary real logarithm to the base e, r>0, and the angle t is chosen to satisfy -P < t <= P.

Log is then defined for all nonzero complex numbers. We have e^(Log(z))=e^(log(r)+it)=r*e^(it)=z.

Now define z^c=e^(c*Log(z)). (This definition makes (2.718...)^c =e^c because of the way Log is defined.) Then (z^c)^c'=(e^(c*Log(z)))^c'=e^(c'*Log(e^(c*Log(z)))=e^(c'c*Log(z))=z^(c'c).

Restricting the imaginary part of the complex log to some other half open interval of length 2P will also give (z^c)^c'=z^(c'c) with the similar definition of z^c, but z^c may then have a different value. The angle assigned to z is what makes the difference. The angle of 1 may be taken to be any one of the values 2nP so that 1^i can be defined to have any of the values e^(i*2inP)=e^(-2nP). This does not mean that e^0=e^(-2P)=e^(-4P) etc.

Edited on November 29, 2003, 8:01 pm

Posted by Richard on 2003-11-27 18:58:30