Chapter 6.8, Problem 50WE

To determine

To show that the conjugates $\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\frac{1}{2}-\frac{\sqrt{3}}{2}i$ are also reciprocals

Explanation of Solution

  $\frac{1}{2}+\frac{\sqrt{3}}{2}i=\frac{1}{2}\left(1+\sqrt{3}i\right)$

Now we simplify the reciprocal of $\frac{1}{2}\left(1+\sqrt{3}i\right)$

We get,

  $\frac{1}{\frac{1}{2}\left(1+\sqrt{3}i\right)}=\frac{2}{\left(1+\sqrt{3}i\right)}$

To solve this we have to multiply numerator and denominator by the conjugate of the denominator.

Here we shall also use the identity $\left(a+b\right)\left(a-b\right)=a^2-b^2$

  $\begin{array}{c}\frac{2}{\left(1+\sqrt{3}i\right)}\times \frac{1-\sqrt{3}i}{1-\sqrt{3}i}=\frac{2\left(1-\sqrt{3}i\right)}{{\left(1\right)}^2-{\left(\sqrt{3}i\right)}^2}\\ =\frac{2-2\sqrt{3}i}{1-3i^2}\\ =\frac{2-2\sqrt{3}i}{1+3}\\ =\frac{2-2\sqrt{3}i}{4}\\ =\frac{1}{2}-\frac{\sqrt{3}}{2}i\end{array}$

Therefore, we can confirm that the conjugates $\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\frac{1}{2}-\frac{\sqrt{3}}{2}i$ are also reciprocals