Chapter 6.6, Problem 142E

To determine

To explain:

The error in the following calculation:

  $\frac{20\left(\cos 90°+i\sin 90°\right)}{5\left(\cos 60°+i\sin 60°\right)}=2\sqrt{3}-2i$

Answer to Problem 142E

The error is right hand side calculated by the division of two complex numbers $20\left(\cos 90°+i\sin 90°\right)$ and $5\left(\cos 60°+i\sin 60°\right)$ is $2\sqrt{3}-2i$ .

It should be $2\sqrt{3}+2i$

Explanation of Solution

Given information:

  $\frac{20\left(\cos 90°+i\sin 90°\right)}{5\left(\cos 60°+i\sin 60°\right)}$

Formula used:

The division rule for two complex numbers:

  $\begin{array}{l}\frac{z_1}{z_2}=\frac{r_1\left(\cos {\theta }_1+i\sin {\theta }_1\right)}{r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)}\\ \frac{z_1}{z_2}=\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\end{array}$

Calculation:

The division of complex number is calculated as:

  $\frac{20\left(\cos 90°+i\sin 90°\right)}{5\left(\cos 60°+i\sin 60°\right)}$

On comparing it with formula:

  $r_1=20,r_2=5,{\theta }_1=90°,{\theta }_2=60°$

Now,

  $\begin{array}{l}\frac{z_1}{z_2}=\frac{r_1}{r_2}\left[\cos \left({\theta }_1-{\theta }_2\right)+i\sin \left({\theta }_1-{\theta }_2\right)\right]\left[0,1\right]\\ \frac{20\left(\cos 90°+i\sin 90°\right)}{5\left(\cos 60°+i\sin 60°\right)}=\frac{20}{5}\left[\cos \left(90°-60°\right)+i\sin \left(90°-60°\right)\right]\\ \frac{20\left(\cos 90°+i\sin 90°\right)}{5\left(\cos 60°+i\sin 60°\right)}=4\left[\cos \left(30\right)°+i\sin \left(30\right)°\right]\\ \frac{20\left(\cos 90°+i\sin 90°\right)}{5\left(\cos 60°+i\sin 60°\right)}=4\left[\frac{\sqrt{3}}{2}+\frac{1}{2}i\right]\\ \frac{20\left(\cos 90°+i\sin 90°\right)}{5\left(\cos 60°+i\sin 60°\right)}=2\sqrt{3}+2i\end{array}$

So, in the given calculation of the problem the right hand side is $2\sqrt{3}-2i$ is an error because by above calculation the right hand side should be $2\sqrt{3}+2i$