[tex]\displaystyle(3t-2i)^5=\binom50(3t)^5(-2i)^0+\binom51(3t)^4(-2i)^1+\binom52(3t)^3(-2i)^2+\binom53(3t)^2(-2i)^3+\binom54(3t)^1(-2i)^4+\binom55(3t)^0(-2i)^5[/tex]
where [tex]\dbinom50,\dbinom51,\ldots[/tex] are the numbers in the fifth row of Pascal's triangle. They're given by
[tex]\dbinom5n=\dfrac{5!}{n!(5-n)!}[/tex]
So the expansion is
[tex]\displaystyle(3t-2i)^5=243t^5-810it^4-1080t^3+720it^2+240t-32i[/tex]
