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Hi,

Here's my implementation of z' = z^5 + c:
   http://www.vanheusden.com/fractals/mand_z5+c_4.0.png


The source-code van be found here: http://www.vanheusden.com/fractals/

(hopefully I did not make any mistakes while expanding that formula to real/img part)

When you say z^5+c, which number format do you mean ?
If ordinary complex then Trifox is correct - it's definitely not correct.

To expand a complex power to reals 1st start with Pascal's triangle to the row required for your power, e.g. for z^5 start with:

So for z^5 that's   "1, 5, 10, 10, 5, 1"
Now add your x/y terms starting with the highest power of x and y^0 (i.e. 1) on the left decrementing the x power and incrementing the y power across the row - remembering that for each y we also have the equivalent power of i:

    x^5, 5*x^4*y*i, 10*x^3*y^2*i^2, 10*x^2*y^3*i^3, 5*x*y^4*i^4, y^5*i^5

Now consider the result of raising i to the power (i.e. giving i, -1, -i, or +1):

   x^5, 5*x^4*y*i,  -10*x^3*y^2, -10*x^2*y^3*i, 5*x*y^4, y^5*i

giving z^5 = x^5 - 10*x^3*y^2 + 5*x*y^4 + i*(5*x^4*y - 10*x^2*y^3 + y^5)

i.e. the Mandy iteration done longhand in Ultra Fractal terms:

 x = real(z)
 y = imag(z)
 z = x^5 - 10*x^3*y^2 + 5*x*y^4 + flip(5*x^4*y - 10*x^2*y^3 + y^5) + #pixel

Edit: Just to add that for higher powers there are formulas for getting the coefficients for a given row of Pascal's triangle - see Wikipedia.