Question 1074556
<pre>
{{{(z-i)^4=i}}}

First find the 4 fourth roots of i, which are

{{{1(cos(pi/2+2pi*n)^""+i*sin(pi/2+2pi*n))}}}, for n=0,1,2,3

{{{cos(((1+4n)/2)^""pi)+i*sin(((1+4n)/2)^""pi)}}}. for n=0,1,2,3

Then taking 4th roots of both sides we have:

{{{z-1=cos(((1+4n)/2)^""pi)+i*sin(((1+4n)/2)^""pi)}}}, for n=0,1,2,3

So

{{{z=1+cos(((1+4n)/2)^""pi)+i*sin(((1+4n)/2)^""pi)}}}, for n=0,1,2,3 

Now substitute n=0,1,2, and 3 and simplify and you'll have the 
4 solutions.

Edwin</pre>