Question

Conjugate 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.

$z-\overline{z}\text{isapureimaginarynumber.}$

Short answer

It is shown that $z-\overline{z}$is a pure imaginary 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 addition of complex numbers.

To add or subtract two complex numbers, simply add/subtract the real parts together and the imaginary parts together. That is,

$\left(a+bi\right)+\left(c+di\right)=\left(a+c\right)+\left(b+d\right)i$

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

Step 3. Evaluate the expressions.

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

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

Thus, the difference$z-\overline{z}$ is a pure imaginary number.