# How do you simplify 2(cos(pi/2)isin(pi/2))*2(cos((3pi)/2)+isin((3pi)/2)) and express the result in rectangular form?

Sep 11, 2017

#### Explanation:

I prefer the $c i s$ notation for polar form, so if we write it like that, we get, calling the above expression $z \cdot w$, this:

$z \cdot w = \left(2 \cdot c i s \left(\frac{\pi}{2}\right)\right) \cdot \left(2 \cdot c i s \left(\frac{3 \pi}{2}\right)\right)$

The product rule for polar form states that for two numbers, ${z}_{1} = {r}_{1} \cdot c i s \left({\theta}_{1}\right)$ and ${z}_{2} = {r}_{2} \cdot c i s \left({\theta}_{2}\right)$, you have ${z}_{1} \cdot {z}_{2} = {r}_{1} \cdot {r}_{2} \cdot c i s \left({\theta}_{1} + {\theta}_{2}\right)$.

Doing this process, we get the following:

$z \cdot w = 4 \cdot c i s \left(2 \pi\right) \equiv 4 \cdot c i s \left(0\right)$.

Since $c i s \left(0\right) = \cos \left(0\right) + i \cdot \sin \left(0\right)$, we get the final result:

$z \cdot w = 4$.