# Math Help - Find the Fifth Roots of the Complex Number

1. ## Find the Fifth Roots of the Complex Number

Hello first time poster here! Just found out about this forum. I have a fairly simple question for most of you all I think, so here goes.

This question has several parts, I'm currently stuck on the first.

Find the fifth roots of the complex number.
98.6(cos(5.06)+sin(5.06)i)
Where theta=5.06 is in radians. There are five roots in the interval [0,2TT) and they all have the same r>0

1. Find r>0
2. Find the first angle
3. Find the second angle
4. Find the third angle
5. Find the forth angle
6. Find the fifth angle

So I'm stuck on number one and I thought I knew how to do these. What I got is:

98.6(cos(5.06)+sin(5.06)i) = 33.58835505 - 92.70265587i
r = sqrt((33.588)^2+(-92.703)^2)

Thanks!

2. Originally Posted by beanus
Hello first time poster here! Just found out about this forum. I have a fairly simple question for most of you all I think, so here goes.

This question has several parts, I'm currently stuck on the first.

Find the fifth roots of the complex number.
98.6(cos(5.06)+sin(5.06)i)
Where theta=5.06 is in radians. There are five roots in the interval [0,2TT) and they all have the same r>0

1. Find r>0
2. Find the first angle
3. Find the second angle
4. Find the third angle
5. Find the forth angle
6. Find the fifth angle

So I'm stuck on number one and I thought I knew how to do these. What I got is:

98.6(cos(5.06)+sin(5.06)i) = 33.58835505 - 92.70265587i
r = sqrt((33.588)^2+(-92.703)^2)

Thanks!

You could have written, $r = \sqrt{(98.6(\cos(5.06))^2+(98.6\sin(5.06))^2}$, since that's where the 33.588 and -92.703 come from.

But $(98.6(\cos(5.06))^2+(98.6\sin(5.06))^2 = (98.6)^2\left[\cos^2(5.06)+\sin^2(5.06)\right]$

That should help with #1.

3. The online math program says that answer is wrong too.

4. $\displaystyle w_k=r^{\frac{1}{n}}\left[cos\left(\frac{\theta+2\pi k}{n}\right)+\mathbf{i}sin\left(\frac{\theta+2\pi k}{n}\right)\right]$

$\displaystyle k,n\in\mathbb{Z} \ \mbox{and} \ k=0,1,2,3,4 \ \mbox{and} \ n=5$

5. Originally Posted by beanus
Hello first time poster here! Just found out about this forum. I have a fairly simple question for most of you all I think, so here goes.

This question has several parts, I'm currently stuck on the first.

Find the fifth roots of the complex number.
98.6(cos(5.06)+sin(5.06)i)
Where theta=5.06 is in radians. There are five roots in the interval [0,2TT) and they all have the same r>0

1. Find r>0
2. Find the first angle
3. Find the second angle
4. Find the third angle
5. Find the forth angle
6. Find the fifth angle

So I'm stuck on number one and I thought I knew how to do these. What I got is:

98.6(cos(5.06)+sin(5.06)i) = 33.58835505 - 92.70265587i
r = sqrt((33.588)^2+(-92.703)^2)

Thanks!
You've calculated $|z|$ for the complex number.

The fifth roots have $\displaystyle\ r=|z|^{\frac{1}{5}$ as dwsmith has shown.

6. Any problems to do with exponents are easiest to solve in exponential form $\displaystyle r\,e^{i\theta}$.

In your case, $\displaystyle 98.6[\cos{(5.06)}+ i\sin{(5.06)}] = 98.6e^{5.06i}$.

So the first fifth root is $\displaystyle \left(98.6e^{5.06i}\right)^{\frac{1}{5}} = 98.6^{\frac{1}{5}}e^{1.012i}$.

The rest are evenly spaced around a circle, so all have the same magnitude and are separated by an angle of $\displaystyle \frac{2\pi}{5}$. Can you find the rest?

7. Originally Posted by beanus
Hello first time poster here! Just found out about this forum. I have a fairly simple question for most of you all I think, so here goes.

This question has several parts, I'm currently stuck on the first.

Find the fifth roots of the complex number.
98.6(cos(5.06)+sin(5.06)i)
Where theta=5.06 is in radians. There are five roots in the interval [0,2TT) and they all have the same r>0

1. Find r>0
2. Find the first angle
3. Find the second angle
4. Find the third angle
5. Find the forth angle
6. Find the fifth angle

So I'm stuck on number one and I thought I knew how to do these. What I got is:

98.6(cos(5.06)+sin(5.06)i) = 33.58835505 - 92.70265587i
r = sqrt((33.588)^2+(-92.703)^2)
The point everyone has been trying to make is that if a complex number is written in the form " $a(cos(\theta)+ i sin(\theta))$", then r= a. Here, r= 98.6 without doing any calculation at all.

Thanks!

8. I tried entering r=98.6 and the online program still says that is wrong. At this point I'm starting to think that the online program is wrong.

9. I am sorry for anyone doing an online mathematics program.
They are notoriously bad about not excepting anything slightly different from what is expected.

The truth is $r = \sqrt[5]{{98.6}}=2.505$. I don’t know what form is expected.

The argument $\varphi = \frac{{5.06}}{5} = 1.012$.

The additive factor is $\[\frac{{2\pi }}{5} = 1.257$ , thus the next angle should be $\varphi_2=2.269$.

10. Plato, that was the answer the program was looking for. I ended up getting it right on the last attempt before it counted the homework question wrong. Thanks!

Thank you all for the help!

11. -

12. Originally Posted by beanus
I tried entering r=98.6 and the online program still says that is wrong. At this point I'm starting to think that the online program is wrong.
When the problem say "Find r> 0" it means the "r" for the fifth root, not the given number! As I, and others, have said, the "r" for this given number is 98.6 and several people have told you that "r" for the fifth root is $\sqrt[5]{98.6}= 98.6^{1/5}$, the fifth root of 98.6.