What is the area of this polygon? Answer in units
1 answer:
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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
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<span>∴ The modulus of the given complex number = l z l = 81
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∴ The modulus of the fourth root =![\sqrt[4]{z} = \sqrt[4]{81} = 3](https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7Bz%7D%20%3D%20%20%5Csqrt%5B4%5D%7B81%7D%20%3D%203)
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 =
</span>
If we have the complex number ⇒ z = a ( cos θ + i sin θ)
∴
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 =
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 =
Step-by-step explanation:
Angles in the same segment. The angles at the circumference subtended by the same arc are equal. More simply, angles in the same segment are equal.
Picture 1 a° = a° they are the same
Picture 2 p= 52° q= 40° Angles in the same segment are equal. if they ask you to calculate you just show both angles p= 52 and q=40.
Picture 3 + 4 Let the obtuse angle MOQ = 2x.
Using the circle theorem, the angle at the centre is twice the angle at the circumference.
Therefore picture 4 tells us and proves;
Angle MNQ = x and angle MPQ = x.
Picture 5
A cyclic quadrilateral is a quadrilateral drawn inside a circle. Every corner of the quadrilateral must touch the circumference of the circle.
The second shape is not a cyclic quadrilateral. One corner does not touch the circumference.
Picture 6
The opposite angles in a cyclic quadrilateral add up to 180°.
a + c = 180°
b + d = 180°
Picture 7
Example
Calculate the angles a and b.
The opposite angles in a cyclic quadrilateral add up to 180°.
This is a picture that couldn't be added but has a quadrilateral occupying half the triangle consisting of 3 arcs the middle one was 140 and proved part of one triangle. The x (a) missing angle showed below a = 180-60 = 120 °
So hopefully this will help you remember the format here for quadrilateral shapes within a circle.
b = 180 - 140 = 40°
a = 180 - 60 = 120°
