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.

${\left(\overline{z}\right)}^2=\overline{z^2}$

Short answer

It is shown that ${\left(\overline{z}\right)}^2=\overline{z^2}$.

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, we will solve the left hand side expression and right hand expression and show that both are equal.

To find the value of the left hand side expression, first find the complex conjugate of z and then expand the exponent as shown below

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

Now, to solve the right hand side expression, first expand the exponent and then find the complex conjugate of resulting as shown below

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

Since the left hand side expression is equal to the right hand side expression, so we can conclude that ${\left(\overline{z}\right)}^2=\overline{z^2}$.