Question

Let z =38 (cos(pi/8)+ i sin(pi/8)) and w = 2 (cos( pi/16 ) + i sin(pi/16))

What is the product of zw?

Answer 1

Answer:

$zw=76 * (cos(\frac{3\pi}{16}) +isin(\frac{3\pi}{16}) )$

Step-by-step explanation:

Given

$z =38 (cos(\frac{\pi}{8})+ i sin(\frac{\pi}{8})$ and

$w = 2 (cos(\frac{\pi}{16} ) + i sin(\frac{\pi}{16})$

Required

Determine zw

Given that:

$x = v_1(cos(a) + isin(a))$ and $y = v_2(cos(b) + isin(b))$

$x * y = (v_1 * v_2)*((cos(a+b) + isin(a+b))$

So, we have:

$zw=(38 * 2) * (cos(\frac{\pi}{8} + \frac{\pi}{16}) + isin(\frac{\pi}{8} + \frac{\pi}{16}))$

$zw=76 * (cos(\frac{2\pi + \pi}{16})+isin(\frac{2\pi + \pi}{16}))$

$zw=76 * (cos(\frac{3\pi}{16}) +isin(\frac{3\pi}{16}) )$

Answer 2

The value of ZW is  $\begin{aligned} (76) [Cos(\frac{3\pi}{16})+ \i\ Sin(\frac{3\pi}{16})]\end{aligned}$.

Given to us,

$\begin{aligned} Z &= 38 [cos(\frac{\pi}{8}) + i\ sin(\frac{\pi}{8}) ]\end{aligned}$,

$\begin{aligned}W &= 2 [cos(\frac{\pi}{16}) + i\ sin(\frac{\pi}{16}) ]\end{aligned}$,

We know that,

${x = v_1 [ cos\ (a) +i\ sin\ (a)]}$  and  ${y = v_2 [ cos(b) +i\ sin(b)]}$

So,

$xy = (v_1 \times v_2) [ cos(a+b) + i\ sin(a+b)]$

Therefore, we can write it as,

$\begin{aligned}ZW &= [38cos(\frac{\pi}{8}) + 38\ i\ sin(\frac{\pi}{8}) ]\times [2cos(\frac{\pi}{16}) + 2\ i\ sin(\frac{\pi}{16}) ]\\&= (38 \times 2) \{Cos[\frac{\pi}{8} + \frac{\pi}{16}]\}+ \{i\ Sin[\frac{\pi}{8} + \frac{\pi}{16}]\} \end{aligned}$

        $\begin{aligned}&= (76) [Cos(\frac{3\pi}{16})+ \i\ Sin(\frac{3\pi}{16})]\end{aligned}$

Hence, the value of ZW is  $\begin{aligned} (76) [Cos(\frac{3\pi}{16})+ \i\ Sin(\frac{3\pi}{16})]\end{aligned}$.

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