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.
[tex]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)\\\\[/tex]
We can abbreviate them into the shorthand form
[tex]z_1 = r_1*\text{cis}(\theta_1)\\\\z_2 = r_2*\text{cis}(\theta_2)\\\\[/tex]
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:
[tex]z_1*z_2 = (r_1*r_2)*\text{cis}(\theta_1+\theta_2)[/tex]
We do two basic things:
- Multiply the r values out front
- 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
[tex]z_1 = 5*\left(\cos(15^{\circ})+i*\sin(15^{\circ})\right)\\\\z_2 = 3*\left(\cos(70^{\circ})+i*\sin(70^{\circ})\right)\\\\[/tex]
abbreviate into
[tex]z_1 = 5*\text{cis}(15^{\circ})\\\\z_2 = 3*\text{cis}(70^{\circ})\\\\[/tex]
then those multiply to
[tex]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)\\\\[/tex]
which is why choice D is the final answer.