Answer:
First Attachment : Option A,
Second Attachment : Option C
Step-by-step explanation:
We are given that,
z₁ = [tex]3(\cos ((\pi )/(6))+i\sin ((\pi )/(6)))[/tex] and z₂ = [tex]4(\cos ((\pi )/(3))+i\sin ((\pi )/(3)))[/tex]
Therefore if we want to determine z₁( z₂ ), we would have to find the trigonometric form of the following expression,
[tex]3(\cos ((\pi )/(6))+i\sin ((\pi )/(6)))*4(\cos ((\pi )/(3))+i\sin ((\pi )/(3)))[/tex]
( Combine expressions )
= [tex]12(\cos ( \pi /6+\pi / 3 ) + i\sin (\pi /6 +\pi / 3 )[/tex]
( Let's now add [tex]\pi / 6 + \pi / 3[/tex], further simplifying this expression )
[tex]\frac{\pi }{6}+\frac{\pi }{3} = \frac{\pi }{6}+\frac{\pi 2}{6} = \frac{\pi +\pi 2}{6} = \frac{3\pi }{6} = \pi / 2[/tex]
( Substitute )
[tex]12(\cos ( \pi /2 ) + i\sin ( \pi /2 ) )[/tex]
And therefore the correct solution would be option a, for the first attachment.
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For this second attachment, we would have to solve for the following expression,
[tex]\frac{3\left(\cos \left(\frac{\pi \:}{6}\right)+i\sin \left(\frac{\pi \:}{6}\right)\right)}{4\left(\cos \left(\frac{\pi \:}{3}\right)+i\sin \left(\frac{\pi \:}{3}\right)\right)}[/tex]
From which we know that cos(π/6) = √3 / 2, sin(π/6) = 1 / 2, cos(π/3) = 1 / 2, and sin(π/3) = √3 / 2. Therefore,
[tex]\:\frac{3\left(\cos \left(\frac{\pi }{6}\right)+i\sin \left(\frac{\pi }{6}\right)\right)}{4\left(\cos \left(\frac{\pi }{3}\right)+i\sin \left(\frac{\pi }{3}\right)\right)}:\quad \frac{3\sqrt{3}}{8}-i\frac{3}{8}[/tex]
[tex]\frac{3\sqrt{3}}{8}-i\frac{3}{8} = \frac{3\sqrt{3}}{8}-\frac{3}{8}i[/tex]
Our solution for the second attachment will be option c.
