Question

HELP PLS!! ILL GIVE POINTS !! :(( m

Answer 1

Answer:  Choice D

15(cos85° + i sin85°)

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Explanation:

Let's say we had these two general complex numbers, which are in polar form.

$z_1 = r_1*\left(\cos(\theta_1)+i*\sin(\theta_1)\right)\\\\z_2 = r_2*\left(\cos(\theta_2)+i*\sin(\theta_2)\right)$

We can abbreviate them into the shorthand form

$z_1 = r_1*\text{cis}(\theta_1)\\\\z_2 = r_2*\text{cis}(\theta_2)$

The notation "cis" stands for "cosine i sine".

Now that we have those complex numbers set up, multiplying them is as simple as saying this:

$z_1*z_2 = (r_1*r_2)*\text{cis}(\theta_1+\theta_2)$

We do two basic things:

1. Multiply the r values out front
2. Add the theta values inside the the cis function

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With all that in mind, let's tackle the problem your teacher gave you.

The given complex numbers

$z_1 = 5*\left(\cos(15^{\circ})+i*\sin(15^{\circ})\right)\\\\z_2 = 3*\left(\cos(70^{\circ})+i*\sin(70^{\circ})\right)$

abbreviate into

$z_1 = 5*\text{cis}(15^{\circ})\\\\z_2 = 3*\text{cis}(70^{\circ})$

then those multiply to

$z_1*z_2 = (r_1*r_2)*\text{cis}(\theta_1+\theta_2)\\\\z_1*z_2 = (5*3)*\text{cis}(15+70)\\\\z_1*z_2 = 15\text{cis}(85^{\circ})\\\\z_1*z_2 = 15\left(\cos(85^{\circ})+i\sin(85^{\circ})\right)$

which is why choice D is the final answer.