Chapter 2.4, Problem 47E

To determine

To write the complex conjugate of $\sqrt{-29}$ and then multiply it with its complex conjugate.

Answer to Problem 47E

Complex conjugate: $-\sqrt{29}i$

Product of the number and its conjugate: 29

Explanation of Solution

Given:

The complex number $\sqrt{-29}$ .

Concept used:

To write a complex conjugate of a complex number, reverse the sign of imaginary part.

The unit of imaginary number, $i=\sqrt{-1}$ .And, $i^2=-1$ .

The standard form a complex number is given as $a+bi$ , where a and b are real numbers.

Calculation:

First write the given complex number in standard form follows:

  $\begin{array}{c}\sqrt{-29}=\sqrt{-1}\sqrt{29}\\ =i\sqrt{29}\\ =0+\sqrt{29}i\end{array}$

Reverse the sign of the imaginary part of the complex number to write its complex conjugate. So, the complex conjugate of $\sqrt{-29}$ is $-\sqrt{29}i$ .

Now, multiply the number and its complex conjugate as follows:

  $\begin{array}{c}\left(0+\sqrt{29}i\right)\left(0-\sqrt{29}i\right)=0^2+{\left(\sqrt{29}\right)}^2\left(\begin{array}{l}\text{Apply the rule: }\\ \left(a+bi\right)\left(a-bi\right)=a^2+b^2\end{array}\right)\\ =0+29\\ =29\end{array}$

Conclusion:

The complex conjugate of the given number is $-\sqrt{29}i$ and the product of number and its complex conjugate is 29.