Chapter 6.6, Problem 75E

To determine

To calculate:

The product of following complex number in the complex plane:

  $\left[2\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\right]\left[\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)\right]$

Answer to Problem 75E

The product of $\left[2\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\right]\left[\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)\right]$ is $-1$ .

Explanation of Solution

Given information:

  $\left[2\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\right]\left[\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)\right]$

Formula used:

The product rule for two complex numbers:

  $\begin{array}{l}{z}_1{z}_2=r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)\cdot r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)\\ z_1{z}_2=r_1\cdot r_2\left[\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right]\end{array}$

Calculation:

The multiplication of complex number is calculated as:

  $\left[2\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\right]\left[\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)\right]$

On comparing it with formula:

  $r_1=2,r_2=\frac{1}{2},{\theta }_1=\frac{2\pi }{3},{\theta }_2=\frac{\pi }{3}$

Now,

  $\begin{array}{l}\left[2\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\right]\left[\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)\right]=2\cdot \frac{1}{2}\left[\cos \left(\frac{2\pi }{3}+\frac{\pi }{3}\right)+i\sin \cos \left(\frac{2\pi }{3}+\frac{\pi }{3}\right)\right]\\ \left[2\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\right]\left[\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)\right]=1\left[\cos \left(\frac{3\pi }{3}\right)+i\sin \left(\frac{3\pi }{3}\right)\right]\\ \left[2\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\right]\left[\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)\right]=\cos \pi +i\sin \pi \\ \left[2\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\right]\left[\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)\right]=-1+i\left(0\right)\\ \left[2\left(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\right)\right]\left[\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right)\right]=-1\end{array}$

Therefore, the product of given two complex numbers in complex plane is $\left(-1,0\right)$ on real axis.

So, the product complex number is $-1$