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IrinaK [193]
3 years ago
6

Find the complex fourth roots of 81(cos(3pi/8) + i sin(3pi/8))

Mathematics
1 answer:
BartSMP [9]3 years ago
7 0
By using <span>De Moivre's theorem:
</span>
If we have the complex number ⇒ z = a ( cos θ + i sin θ)
∴ \sqrt[n]{z} =  \sqrt[n]{a} \ (cos \  \frac{\theta + 360K}{n} + i \ sin \ \frac{\theta +360k}{n} )
k= 0, 1 , 2, ..... , (n-1)


For The given complex number <span>⇒ z = 81(cos(3π/8) + i sin(3π/8))
</span>

Part (A) <span>find the modulus for all of the fourth roots
</span>
<span>∴ The modulus of the given complex number = l z l = 81
</span>
∴ The modulus of the fourth root = \sqrt[4]{z} =  \sqrt[4]{81} = 3

Part (b) find the angle for each of the four roots

The angle of the given complex number = \frac{3 \pi}{8}
There is four roots and the angle between each root = \frac{2 \pi}{4} =  \frac{\pi}{2}
The angle of the first root = \frac{ \frac{3 \pi}{8} }{4} =  \frac{3 \pi}{32}
The angle of the second root = \frac{3\pi}{32} +  \frac{\pi}{2} =  \frac{19\pi}{32}
The angle of the third root = \frac{19\pi}{32} +  \frac{\pi}{2} =  \frac{35\pi}{32}
The angle of the  fourth root = \frac{35\pi}{32} +  \frac{\pi}{2} =  \frac{51\pi}{32}

Part (C): find all of the fourth roots of this

The first root = z_{1} = 3 ( cos \  \frac{3\pi}{32} + i \ sin \ \frac{3\pi}{32})
The second root = z_{2} = 3 ( cos \  \frac{19\pi}{32} + i \ sin \ \frac{19\pi}{32})

The third root = z_{3} = 3 ( cos \  \frac{35\pi}{32} + i \ sin \ \frac{35\pi}{32})
The fourth root = z_{4} = 3 ( cos \  \frac{51\pi}{32} + i \ sin \ \frac{51\pi}{32})
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1. A student took 60 minutes to answer a combination of 20 multiple-choice and extended-response questions. She took 2 minutes t
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Answer:

1) m=15 and r =5

2) 4 and 2 ml

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Step-by-step explanation:

given that m multiple choice questions and r extended response questions.

Also given that a) m+r = total no of questions =20 ... i and

                        2m+6r = total time taken = 60  ... ii

b) Divide second equation by 2, m+3r = 30  ... iii

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Subtract to get 2r =10 or r =5

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--------------------------------------

2) Let a litres of 20% solution and b litres of 50% solution be mixed

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Resulting solution = 30% of 6 ml = 1.8 = 0.2a+0.5b ... ii

b) Solve i and ii

b = 6-a: substitute in ii.

0.2a + 3-0.5a = 1.8

Or a = 4 ml and b = 2ml

Verify: Total volume = a+b =6ml and

concentration = 0.2(4)+0.5(2) = 1.8 = 30% of 6 ml.

Thus verified

---------------------------------

3) 2x-3y=4 ...i

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Because x term has the same coefficient in both the equations, elimination is easier.

i-ii gives 2y =-2 or y =-1

Substitute in i, 2x+3 =4 or x = 0.5

So answer is x = 0.5 and y =1


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