Question

Recall that the symbol $\overline{z}$represents the complex conjugate of z. If $z=a+bi\text{and}w=c+di$, show that each statement is true.

$z\cdot \overline{z}\text{isarealnumber.}$

Short answer

It is shown that $z\cdot \overline{z}$is a real number.

Step-by-step solution

Step 1. Concept of complex conjugate.

The complex conjugate of the complex number$a+bi$ is represented by a horizontal bar over it, that is, the complex conjugate of$a+bi$ is denoted by $\overline{a+bi}$.

The complex conjugate of any complex number can be obtained by changing the sign of the imaginary part of the complex number. The complex conjugate of $a+bi$is $a-bi$.

The product of a complex number and its complex conjugate is given by

$\left(a+bi\right)\left(a-bi\right)=a^2+b^2$

The imaginary number i is defined as the square root of negative one. That is, $i^2=-1$.

Step 2. Concept of multiplication of complex numbers.

To multiply two complex numbers, multiply them as binomial expressions and use the result of imaginary number, $i^2=-1$.

The FOIL method of multiplying binomials is

$\left(a+b\right)+\left(c+d\right)=a\cdot c+a\cdot d+b\cdot c+b\cdot d$

The product of a complex number and its complex conjugate is given by

$\left(a+bi\right)\left(a-bi\right)=a^2+b^2$

Step 3. Evaluate the expressions.

In order to show that the given statement is true, find the complex conjugate of z and multiply it with the complex number z as shown below

$$\begin{gathered} z\cdot \overline{z}=\left(a+bi\right)\cdot \left(\overline{a+bi}\right) \\ =\left(a+bi\right)\cdot \left(a-bi\right) \\ =a\cdot a+a\cdot \left(-bi\right)+bi\cdot a+bi\cdot \left(-bi\right) \\ =a^2-abi+abi-b^2{i}^2 \\ =a^2+\left(-ab+ab\right)i-b^2\left(-1\right) \\ =a^2+\left(0\right)i+b^2 \\ =a^2+b^2 \end{gathered}$$

Thus, the product$z\cdot \overline{z}$ is a real number.