By using <span>De Moivre's theorem:
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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} )](https://tex.z-dn.net/?f=%20%5Csqrt%5Bn%5D%7Bz%7D%20%3D%20%20%5Csqrt%5Bn%5D%7Ba%7D%20%5C%20%28cos%20%5C%20%20%5Cfrac%7B%5Ctheta%20%2B%20360K%7D%7Bn%7D%20%2B%20i%20%5C%20sin%20%5C%20%5Cfrac%7B%5Ctheta%20%2B360k%7D%7Bn%7D%20%29)
k= 0, 1 , 2, ..... , (n-1)
For The given complex number <span>⇒ z = 81(cos(3π/8) + i sin(3π/8))
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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
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∴ The modulus of the fourth root =
Part (b) find the angle for each of the four roots
The angle of the given complex number =

There is four roots and the angle between each root =

The angle of the first root =

The angle of the second root =

The angle of the third root =

The angle of the fourth root =
Part (C): find all of the fourth roots of this
The first root =

The second root =

The third root =

The fourth root =
ANSWER

EXPLANATION
We use the sine rule for solving triangles.
This is given by the formula,

From the triangle,

We multiply both sides of the equation by 60 to get,


We solve for x to obtain,


To the nearest tenth, we round to one decimal place to get,
Answer:
b = 12.6
Step-by-step explanation:
a*2 + b*2 = c*2
6*2 + b*2 = 14*2
36 + b*2 = 196
Subtract 36 from both sides
36 + b*2 = 196
-36 -36
b*2 = 160
Find the square root of both
b = 12.64911064 or 12.6
Answer: A
Step-by-step explanation: