You need to use the binomial formula, such that:
`(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k`
You need to replace 2 for x, 3i for y and 6 for n, such that:
`(2-3i)^6 = 6C0 2^6 + 6C1 2^5*(-3i)^1 + 6C2 2^4*(-3i)^2 + 6C3 2^3*(-3i)^3 + 6C4 2^2*(-3i)^4 + 6C5 2^1*(-3i)^5 + 6C6 (-3i)^6`
By definition, `nC0 = nCn = 1` , hence `6C0 = 6C6 = 1.`
By definition `nC1 = nC(n-1) = n` , hence `6C1 = 6C5 = 6.`
By definition `nC2 = nC(n-2) = (n(n-1))/2` , hence `6C2 = 6C4 = (6(6-1))/2 = 15`
`(2-3i)^6 = 2^6 - 6*2^5*3i + 15*2^4*3^2*i^2 - (6!)/(3!*(6-3)!)*2^3*3^3*i^3 + 15*2^2*3^4*i^4 - 6*2*3^5*i^5 + 3^6*i^6`
Using the powers of i yields:
`i = i, i^2 = -1, i^3 = -i, i^4 = 1, i^5 = i, i^6 = -1`
`(2-3i)^6=64-576i-2160+4320i+4860-2916i-729`
`(2-3i)^6=2035+828i`
Hence, expanding the complex number using binomial theorem yields the simplified result `(2-3i)^6 = 2035 + 828i.`
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