How do you simplify 2(cos⁡(π2)isin⁡(π2)⋅2(cos⁡(3π2)+isin⁡(3π2)) and express the result in rectangular form?

I prefer the cis notation for polar form, so if we write it like that, we get, calling the above expression z⋅w, this: z⋅w=(2⋅cis(π2))⋅(2⋅cis(3π2)) The product rule for polar form states that for two numbers, z1=r1⋅cis(θ1)and z2=r2⋅cis(θ2), you have z1⋅z2=r1⋅r2⋅cis(θ1+θ2). Doing this process, we get the following: z⋅w=4⋅cis(2π)≡4⋅cis(0). Since cis(0)=cos⁡(0)+i⋅sin⁡(0), we get the final result: z⋅2=4 Answer: 4.

Explained: "How do you simplify 2(cos⁡(π2)isin⁡(π2)⋅2(cos⁡(3π2)+isin⁡(3π2)) and express the..."

Laney Spears

Answered question

2022-01-29

How do you simplify

2 (\cos (\frac{π}{2}) i \sin (\frac{π}{2}) \cdot 2 (\cos (\frac{3 π}{2}) + i \sin (\frac{3 π}{2}))

and express the result in rectangular form?

Answer & Explanation

Jason Olsen

Beginner2022-01-30Added 14 answers

I prefer the cis notation for polar form, so if we write it like that, we get, calling the above expression

z \cdot w

, this:

z \cdot w = (2 \cdot c i s (\frac{π}{2})) \cdot (2 \cdot c i s (\frac{3 π}{2}))

The product rule for polar form states that for two numbers,

z_{1} = r_{1} \cdot c i s (θ_{1}) and z_{2} = r_{2} \cdot c i s (θ_{2})

, you have

z_{1} \cdot z_{2} = r_{1} \cdot r_{2} \cdot c i s (θ_{1} + θ_{2})

.
Doing this process, we get the following:

z \cdot w = 4 \cdot c i s (2 π) \equiv 4 \cdot c i s (0)

.
Since

c i s (0) = \cos (0) + i \cdot \sin (0)

, we get the final result:

z \cdot 2 = 4

Answer: 4.

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