Question 326119
Convert to {{{r}}}, {{{theta}}} format,
{{{2-2i=2(cos(315)+isin(315))}}}
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{{{z^n=r^n*(cos(n*theta)+i*sin(n*theta))}}}
{{{(2-2i)^5= (sqrt(8))^5*(cos(1575)+i*sin(1575))}}}
{{{(2-2i)^5=(sqrt(8))^5*(cos(135)+i*sin(135))}}}
{{{(2-2i)^5=(sqrt(8))^5*(-(sqrt(2)/2)+(sqrt(2)/2)i)}}}
{{{(2-2i)^5=-128+128i}}}