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If (2 + 3i)^2 (2 - 3i)^2 = a + bi, what are the values of a and b?

a) 4, 12
b) 16, 0
c) 0, 16
d) 12, 4

1 Answer

7 votes

Final answer:

After calculating, the complex numbers multiplication of (2 + 3i)2 and (2 - 3i)2 gives us a real part (a) of 169 and an imaginary part (b) of 0. The answer is not among the provided options, which suggests an error in the question.

Step-by-step explanation:

The student is asking for the resulting real part a and the imaginary part b of the product of two complex numbers raised to the second power, specifically (2 + 3i)2 and (2 - 3i)2. To find the values of a and b, we first expand these expressions:

  • (2 + 3i)2 = 4 + 12i + 9i2
  • (2 - 3i)2 = 4 - 12i + 9i2

Since i2 is equal to -1, we can simplify further:

  • (2 + 3i)2 = 4 + 12i - 9 = -5 + 12i
  • (2 - 3i)2 = 4 - 12i - 9 = -5 - 12i

Multiplying these two expressions, we get:

(-5 + 12i) * (-5 - 12i) = 25 + 60i - 60i + 144 = 25 + 144 = 169

After multiplication, the imaginary terms cancel out, so b is 0, and the real part a is 169. So, the correct answer is (a = 169, b = 0) which means b is not among the provided options, hence it seems to be an error in the options.

User Daniele Faugiana
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