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Chapter 2: Problem 100

Proof Prove that the complex conjugate of the product of two complex numbers \(a_{1}+b_{1} i\) and \(a_{2}+b_{2} i\) is the product of their complex conjugates.

Short Answer

Expert verified
The complex conjugate of the product of two complex numbers is equal to the product of their complex conjugates.

Step by step solution

01

Define the Complex Numbers

First let's define our complex numbers: \(z_{1} = a_{1}+b_{1}i\) and \(z_{2} = a_{2}+b_{2}i\).
02

Find the product of Complex Numbers

Let鈥檚 multiply the two complex numbers: \(z_{1} * z_{2} = (a_{1}+b_{1}i) * (a_{2}+b_{2}i) = a_{1}a_{2} - b_{1}b_{2} + i(a_{1}b_{2} + b_{1}a_{2})\). This is the product \(z_{3}\) of the two complex numbers
03

Find the conjugate of the product

To find the conjugate of the product, all imaginary components change their sign: \(\overline{z_{3}} = a_{1}a_{2} - b_{1}b_{2} - i(a_{1}b_{2} + b_{1}a_{2})\). This is the conjugate of the product of complex numbers
04

Find the conjugates of each complex number

Now find the complex conjugates of \(z_{1}\) and \(z_{2}\): \(\overline{z_{1}} = a_{1} - b_{1}i\) and \(\overline{z_{2}} = a_{2} - b_{2}i\)
05

Find the product of the conjugates

Now multiply the conjugates: \(\overline{z_{1}} * \overline{z_{2}} = (a_{1} - b_{1}i) * (a_{2} - b_{2}i) = a_{1}a_{2} - b_{1}b_{2} - i(a_{1}b_{2} + b_{1}a_{2})\). This is the product of the conjugates of complex numbers
06

Comparing the results

Compare the two results. As we can see both \(\overline{z_{3}}\) and \(\overline{z_{1}} * \overline{z_{2}}\) are equal, so the statement is proved.

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