Which pair of complex factors results in a real-number product?
A. $(8+20 i)(-8-20 i)$
B. $3 i(1-3 i)$
C. $(4+7 i)(4-7 i)$
D. $15(-15 i)$
Answer (2)
Multiply each pair of complex numbers.
Check the imaginary part of the product.
If the imaginary part is zero, the product is a real number.
The pair ( 4 + 7 i ) ( 4 − 7 i ) results in a real number product: ( 4 + 7 i ) ( 4 − 7 i ) .
Explanation
Understanding the Problem We are given four pairs of complex numbers and asked to identify the pair whose product is a real number. A real number has an imaginary part equal to zero.
Calculating the Products Let's calculate the product of each pair:
Pair 1: ( 8 + 20 i ) ( − 8 − 20 i )
Expanding this, we get: 8 ( − 8 ) + 8 ( − 20 i ) + 20 i ( − 8 ) + 20 i ( − 20 i ) = − 64 − 160 i − 160 i − 400 i 2 = − 64 − 320 i + 400 = 336 − 320 i . The imaginary part is -320, so this is not a real number.
Pair 2: 3 i ( 1 − 3 i ) Expanding this, we get: 3 i ( 1 ) + 3 i ( − 3 i ) = 3 i − 9 i 2 = 3 i + 9 = 9 + 3 i . The imaginary part is 3, so this is not a real number.
Pair 3: ( 4 + 7 i ) ( 4 − 7 i ) Expanding this, we get: 4 ( 4 ) + 4 ( − 7 i ) + 7 i ( 4 ) + 7 i ( − 7 i ) = 16 − 28 i + 28 i − 49 i 2 = 16 + 49 = 65 . The imaginary part is 0, so this is a real number.
Pair 4: 15 ( − 15 i ) Expanding this, we get: − 225 i . The imaginary part is -225, so this is not a real number.
Identifying the Real-Number Product From the calculations above, only the product of the third pair, ( 4 + 7 i ) ( 4 − 7 i ) , results in a real number (65).
Final Answer Therefore, the pair of complex factors that results in a real-number product is ( 4 + 7 i ) ( 4 − 7 i ) .
Examples
Complex numbers are used in electrical engineering to analyze alternating current circuits. The impedance of a circuit, which is the opposition to the flow of current, is represented as a complex number. Multiplying complex numbers is essential for calculating power and voltage in AC circuits. For example, if you have a circuit with impedance Z = 4 + 7 i ohms and you want to find a real power value, you might use the conjugate 4 − 7 i in your calculations to eliminate the imaginary components and find the real power dissipated in the circuit.
Only Pair C, ( 4 + 7 i ) ( 4 − 7 i ) , results in a real-number product with a value of 65 . The product has an imaginary part of zero, confirming it is a real number. Therefore, the answer is Pair C.
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