Question 987127
<pre>
{{{z=(1+i)/(1-i)}}}

{{{z}}}{{{""=""}}}{{{(1+i)/(1-i)}}}{{{"×"}}}{{{(1+i)/(1+i)}}}{{{""=""}}}{{{(1+2i+i^2)/(1-i^2)}}}{{{""=""}}}{{{(1+2i+(-1))/(1-(-1))}}}{{{""=""}}}{{{2i/2}}}{{{""=""}}}{{{i}}}

z+z²+z³+z&#8308; = i+i²+i³+i&#8308; = i+(-1)+(i²×i)+(i²×i²) = i-1+(-1×i)+(-1×-1) =

i-1-i+1 = 0

Edwin</pre>