The product of complex conjugates ( a + bi ) ( a − bi ) is a 2 + b 2 .
For ( 6 + 3 i ) ( 6 − 3 i ) , the product is 6 2 + 3 2 = 45 .
For ( 4 − 5 i ) ( 4 + 5 i ) , the product is 4 2 + 5 2 = 41 .
For ( − 3 + 8 i ) ( − 3 − 8 i ) , the product is ( − 3 ) 2 + 8 2 = 73 .
The final answers are: 45 , 41 , 73 .
Explanation
Understanding the Problem We are given three pairs of complex numbers to multiply. Each pair has the form ( a + bi ) ( a − bi ) , where a and b are real numbers, and i is the imaginary unit ( i 2 = − 1 ). This form represents a complex number multiplied by its complex conjugate. The product of a complex number and its conjugate is always a real number and can be calculated as a 2 + b 2 .
Calculating the First Product For the first pair, ( 6 + 3 i ) ( 6 − 3 i ) , we identify a = 6 and b = 3 . The product is a 2 + b 2 = 6 2 + 3 2 = 36 + 9 = 45 .
Calculating the Second Product For the second pair, ( 4 − 5 i ) ( 4 + 5 i ) , we identify a = 4 and b = 5 . The product is a 2 + b 2 = 4 2 + 5 2 = 16 + 25 = 41 .
Calculating the Third Product For the third pair, ( − 3 + 8 i ) ( − 3 − 8 i ) , we identify a = − 3 and b = 8 . The product is a 2 + b 2 = ( − 3 ) 2 + 8 2 = 9 + 64 = 73 .
Final Answer Therefore, the products of the given pairs of complex numbers are 45, 41, and 73, respectively.
Examples
Complex numbers and their conjugates are used extensively in electrical engineering to analyze AC circuits. The impedance of a circuit, which includes resistance and reactance, is often represented as a complex number. Multiplying a complex impedance by its conjugate helps in calculating the power dissipated in the circuit, a crucial aspect of circuit design and analysis. This ensures that electrical systems are efficient and operate within safe parameters.
The products of the given pairs of complex numbers are 45, 41, and 73. Each product is calculated using the formula a 2 + b 2 . This method allows us to find the product of complex numbers and their conjugates effectively.
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