Question 680602
{{{((-2-5i)/3i)(i/i) = (i(-2-5i))/(i(3i))}}}
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i^2 = {{{sqrt(-1)^2 = -1}}}
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{{{(i(-2-5i))/(i(3i)) = (-2i-5i^2)/(3i^2) = (-2i-5(-1))/(3(-1)) = (5-2i)/-3 = -(5-2i)/3 =highlight_green((-5+2i)/3)}}}
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