Chapter 12.1, Problem 61E

To determine

To calculate: The value of $6\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)÷12\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$ . Also express the quotient in rectangular form.

Answer to Problem 61E

The value is $0.2\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)$ . The quotient in rectangular form is $0.46+0.19i$ .

Explanation of Solution

Given information:

The expression $6\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)÷12\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$ .

Calculation:

Consider the expression $6\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)÷12\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)$ .

When two complex numbers in polar form are divided then the radius is simply divided and the argument $\theta$ is ${\theta }_1-{\theta }_2$ .

  $\begin{array}{c}6\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)÷12\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)=\frac{6\left(\cos \frac{5\pi }{8}+i\sin \frac{5\pi }{8}\right)}{12\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)}\\ =0.5\left(\cos \left(\frac{5\pi }{8}-\frac{\pi }{2}\right)+i\sin \left(\frac{5\pi }{8}-\frac{\pi }{2}\right)\right)\\ =0.5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)\end{array}$

Use a calculator to simplify the expression obtained above.

$\begin{array}{c}0.5\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)=0.5\left(0.99+i0.98\right)\\ =0.46+0.19i\end{array}$

Thus, the value is $0.2\left(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8}\right)$ . The quotient in rectangular form is $0.46+0.19i$ .