Question

Find the fifth roots of 32(cos 280° + i sin 280°).

Answer 1

Given your complex number of the form
\[\large z= r\left(\cos \theta \ + \ i \sin \theta\right)\]
If you raise it to an exponent, it's going to look something like this\[\large z^p= r^p\left(\cos \ p\theta \ + \ i \sin \ p\theta\right)\]

Answer 2

That said, if you take the root of a complex number, it's going to look like this:
\[\huge \sqrt[n]z=z^{\frac{1}{n}}= \sqrt[n]r\left(\cos \frac{\theta}{n} \ + \ i \sin \frac{\theta}{n}\right)\]

NOTE: It doesn't end here yet.

Answer 3

So it would go like 32^5(cos (5)280° + i sin (5)280°

Answer 4

Then it becomes 5root32(costheta/5 + isin theta/5)?

Answer 5

Yeah, more like it :)

Answer 6

so it becomes 5root32(cos280/5 + isin 280/5)

Answer 7

Yes. the fifth root of 32 is just 2. You may as well write it like that.
We're not done yet, though, there are more than one fifth roots of this complex number.

Answer 8

ok so 2(cos280/5 + isin280/5).

Answer 9

Simplify 280/5, too...

Answer 10

2(cos56 + isin56)

Answer 11

Okay. Now here's the complication.
\[\large z= r\left(\cos \theta \ + \ i \sin \theta\right)=r\left[\cos \left(\theta+360^o\right)\ + \ i \sin \left(\theta+360^o\right)\right]\]
with theta in degrees, right? Seeings as adding 360 degrees just returns it to its original angle.

Answer 12

So you now also have to apply that rule to
\[\large =r\left[\cos \left(\theta+360^o\right)\ + \ i \sin \left(\theta+360^o\right)\right]\]
\[\large =32\left[\cos \left(280^o+360^o\right)\ + \ i \sin \left(280^o+360^o\right)\right]\]

Answer 13

So the fifth root of 32 is still 2, but now, what's (280+360)/5 ?

Answer 14

128?

Answer 15

@terenzreignz

Answer 16

Use the formula: r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ] where k = 0,1,2,3,4

First 5th root:

k = 0


r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]


(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]


(32)^(1/5)*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]


2*[ cos( (280+360*0)/5 ) + i*sin( (280+360*0)/5 ) ]


2*[ cos( (280+0)/5 ) + i*sin( (280+0)/5 ) ]


2*[ cos( 280/5 ) + i*sin( 280/5 ) ]


2*[ cos( 56 ) + i*sin( 56 ) ]



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Second 5th root:

k = 1


r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]


(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]


(32)^(1/5)*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]


2*[ cos( (280+360*1)/5 ) + i*sin( (280+360*1)/5 ) ]


2*[ cos( (280+360)/5 ) + i*sin( (280+360)/5 ) ]


2*[ cos( 640/5 ) + i*sin( 640/5 ) ]


2*[ cos( 128 ) + i*sin( 128 ) ]


This is the root you found

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Third 5th root:

k = 2


r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]


(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]


(32)^(1/5)*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]


2*[ cos( (280+360*2)/5 ) + i*sin( (280+360*2)/5 ) ]


2*[ cos( (280+720)/5 ) + i*sin( (280+720)/5 ) ]


2*[ cos( 1000/5 ) + i*sin( 1000/5 ) ]


2*[ cos( 200 ) + i*sin( 200 ) ]


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Fourth 5th root:

k = 3


r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]


(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]


(32)^(1/5)*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]


2*[ cos( (280+360*3)/5 ) + i*sin( (280+360*3)/5 ) ]


2*[ cos( (280+1080)/5 ) + i*sin( (280+1080)/5 ) ]


2*[ cos( 1360/5 ) + i*sin( 1360/5 ) ]


2*[ cos( 272 ) + i*sin( 272 ) ]


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Fifth 5th root:

k = 4


r^(1/n)*[ cos( (theta+360*k)/n ) + i*sin( (theta+360*k)/n ) ]


(32)^(1/5)*[ cos( (280+360*k)/5 ) + i*sin( (280+360*k)/5 ) ]


(32)^(1/5)*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]


2*[ cos( (280+360*4)/5 ) + i*sin( (280+360*4)/5 ) ]


2*[ cos( (280+1440)/5 ) + i*sin( (280+1440)/5 ) ]


2*[ cos( 1720/5 ) + i*sin( 1720/5 ) ]


2*[ cos( 344 ) + i*sin( 344 ) ]


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Summary:



Given z = 32 * [ cos( 280 ) + i*sin( 280 ) ], the five 5th roots of z are:


2*[ cos( 56 ) + i*sin( 56 ) ]
2*[ cos( 128 ) + i*sin( 128 ) ]
2*[ cos( 200 ) + i*sin( 200 ) ]
2*[ cos( 272 ) + i*sin( 272 ) ]
2*[ cos( 344 ) + i*sin( 344 ) ]


All angles are in degrees