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Difficulty: MediumComplex Numbers and Operations

Let zz be the complex number resulting from the product (2−3i)(3+i)(2 - 3i)(3 + i), where i=−1i = \sqrt{-1}. If zˉ\bar{z} represents the complex conjugate of zz, what is the value of the product z⋅zˉz \cdot \bar{z}?

  1. A
    32
  2. B
    58
  3. C
    81
  4. D
    95
  5. 130Answer

Answer

130
The correct answer is obtained by first expanding (2−3i)(3+i)(2 - 3i)(3 + i) using FOIL to get 6+2i−9i−3i26 + 2i - 9i - 3i^2. Since i2=−1i^2 = -1, this simplifies to 6−7i−3(−1)=9−7i6 - 7i - 3(-1) = 9 - 7i. The complex conjugate is 9+7i9 + 7i. Multiplying these gives 92−(7i)2=81−49(−1)=1309^2 - (7i)^2 = 81 - 49(-1) = 130.

Step-by-Step Solution

1
Multiply the complex binomials to find zz
z=(2−3i)(3+i)=6+2i−9i−3i2=6−7i−3(−1)=9−7iz = (2 - 3i)(3 + i) = 6 + 2i - 9i - 3i^2 = 6 - 7i - 3(-1) = 9 - 7i
To express the complex number in standard form a+bia + bi, we expand the product using the distributive property and substitute i2=−1i^2 = -1.
2
Determine the complex conjugate of zz, denoted as zˉ\bar{z}
zˉ=9+7i\bar{z} = 9 + 7i
The complex conjugate of a complex number a+bia + bi is a−bia - bi, which is found by reversing the sign of the imaginary part.
3
Calculate the product of zz and zˉ\bar{z}
z⋅zˉ=(9−7i)(9+7i)=92+72=81+49=130z \cdot \bar{z} = (9 - 7i)(9 + 7i) = 9^2 + 7^2 = 81 + 49 = 130
The product of a complex number a+bia + bi and its conjugate a−bia - bi is always a real number equal to a2+b2a^2 + b^2.

Key Concept

Multiplying complex numbers and finding the product of a complex number and its conjugate.
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Topics

Complex Numbers and Operations

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