Expert: Paul Klarreich - 12/13/2008

Question
Hi Paul,
Thanks for your try.
I came up with a solution. Do you think that it is right?
Here it is:
(1+z)^5=-i(1-z)^5 => fifth sqrt((1+z)^5)= fifth sqrt (-i(1-z)^5) => 1+z = -i(1-z) => z=(1+i)/i-1

Thanks again
Stamatis


Answer
Questioner:   Stamatis
Category:  Advanced Math
Private:  No
 
Subject:  Equation with complex numbers
Question:  Hi Paul,
Thanks for your try.
I came up with a solution. Do you think that it is right?
Here it is:
(1+z)^5=-i(1-z)^5 => fifth sqrt((1+z)^5)= fifth sqrt (-i(1-z)^5) => 1+z = -i(1-z) => z=(1+i)/i-1

Thanks again
Stamatis
................
Hi, Stamatis:

I think, when you write fifth sqrt, you mean the fifth root, or  (..)^(1/5)

You may be OK. Looking at your stuff, I got an idea.

............
Here is your equation:

(1+z)^5 + i(1-z)^5=0.

Write:

(1+z)^5 = -i(1-z)^5

Now observe that:
-i = (-1)i  AND  -1 = (-1)^5   AND  i = i^5

The equation is now:

(1+z)^5 = (-1)^5 i^5 (1-z)^5

and now we can take fifth roots on both sides.  (not quite happy with this, but.....)

1+z = (-1) i (1-z)

1+z = -i(1-z)

1+z = -i + iz

z - iz = -1 - i

z(1 - i) = - 1 - i
   - 1 - i
z = ---------
     1 - i

Hmmmmm.  That appears to be identical to your answer.