Question 1135994
.
<pre>
1 - i = cis({{{sqrt(2)}}},{{{7pi/4}}}).


Therefore, by the de Moivre's Theorem


{{{(1-i)^10}}} =cis({{{(sqrt(2))^10}}},{{{70pi/4}}}) = cis({{{2^5}}},{{{6pi/4}}}) = cis(32,{{{3pi/2}}}) = -32*i.    <U>ANSWER</U>
</pre>