How do I find the 5 roots of 1^(1/5) using De Moivre's Theorem?

Do you mean the 5th roots of 1? Or the 5th roots of 1^(1/5), i.e. the 5th roots of the 5th roots of 1?

I assume you mean the former. To begin with,

1 = exp(0°i ) = cos(0°) + i sin(0°)

Take the 5th root, i.e. (1/5)th power of both sides, so that by DeMoivre's theorem,

1^(1/5) = [ cos(0°) + i sin(0°) ]^(1/5)

1^(1/5) = cos((0° + 360°n)/5) + i sin((0° + 360°n)/5)

where n = 0, 1, 2, 3, or 4. Then 1^(1/5) has the 5 possible values,

• cos(0°/5) + i sin(0°/5) = 1

• cos(360°/5) + i sin(360°/5) = cos(72°) + i sin(72°)

• cos(720°/5) + i sin(720°/5) = cos(144°) + i sin(144°)

• cos(1080°/5) + i sin(1080°/5) = cos(216°) + i sin(216°)

• cos(1440°/5) + i sin(1440°/5) = cos(288°) + i sin(288°)

If you wish, or are required to, you can go on to write these in terms of radicals by expanding cos(5x) and sin(5x) (where x = 72° so that 5x = 360°). For example,

cos(5x) = cos⁵(x) - 10 cos³(x) sin²(x) + 5 cos(x) sin⁴(x)

cos(5x) = cos⁵(x) - 10 cos³(x) (1 - cos²(x)) + 5 cos(x) (1 - cos²(x))²

cos(5x) = 16 cos⁵(x) - 20 cos³(x) + 5 cos(x)

which follows from the expansion

(cos(x) + i sin(x))⁵ = cos(5x) + i sin(5x)

due to DeMoivre's theorem and equating the real parts. Then cos(5x) = 1, and you can try to solve for cos(x) :

1 = 16 cos⁵(x) - 20 cos³(x) + 5 cos(x)

Actually, it would be easier to find sin(x) first, since sin(5x) = 0. The expansion in terms of sin(x) will look quite similar to the one shown here.

step-by-step explanation:

30% of $420 is 0.30($420) = $126 and $420 + $126 = $546.

or, 130% of $420 is 1.3($420) = $546.

notice that this is not the $600 original list price with which we started.

why doesn't this work?

the method does not work because 30% of $420 is not the same amount as the 30% of $600. it may seem obvious when stated this way, but it is a very common error made on problems of this type. we cannot just use the same percent and get back to the original amount because we are taking that percent of a different value.

to work problems of this type correctly, it is generally best to write an equation for the situation and then solve.

example: since (100% – 30%)(original price) = (sale price)

70% of the   (original price) is $420.

0.7x   = 420

equation image indicator

x = 600

the original cost was $600.

we may also solve this problem using the two-step method. working the problem in two steps requires us to use more algebra skills. we have to know that x = 1x (identity property of multiplicationtext annotation indicator) and that 1x – 0.3x = 0.7x (subtracting like terms). then we solve the problem as follows:

(original price) – 30%(original price) = sale price

x – 0.3x = 420

0.7x = $420

equation image indicator

x = 600

the original cost was $600.  

note that with both of the above methods we do obtain the correct original amount.

example: after a 5% pay raise, hermione is earning $22,680 per year. what was she earning before the pay raise?

105% of (original salary) = current salary

1.05x = 22,680

equation image indicator

x = 21,600

hermione earned $21,600 per year before the pay raise.

this is an example so maybe you will get how to at least do it!

i leave the calculator work to you.

7. diameter 4 miles, circumference 4π miles

8. radius 0.2 miles, circumference 0.4π miles

[this is a bad problem because the actual radius of a car on the track is less than 0.2 miles so the answer is less than 0.4π miles]

9. radius 6 feet, circumference 12π feet

10. diameter 14 cm, circumference 14π cm

11. diameter 4.8 feet, circumference 4.8π feet