Question

(15-4i)(6-3i) = a +bi

Answer 1

The real part of the result is the product of the real parts of the factors and the product of the imaginary parts:

a = 15*6 - (-4)(-3) = 90 -12 = 78

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You can treat this the same as any other product of binomials except that i² = -1.

... (15 -4i)(6 -3i) = 15(6 -3i) -4i(6 -3i) = 15·6 -15·3i -4·6i +4·3i²

... = 90 -45i -24i +12i² = (90 -12) -69i = 78 -69i