Question

find the 4th roots of the complex number z1=1+sqrt3i

part1: write z1 in polar form
part2: find the mudulus of the root of z1
part3: find the four angles that define the 4th roots of the number z1
part4: what are the fourth roots of z1=sqrt3+i

Answer 1

....

Answer 2

Well, polar form means you need:
\[r(\cos \theta +isin \theta)\]
r is the modulus and theta is the argument.
There are several conversions to go back and forth between rectangular and polar, but we need these two:
\[r = \sqrt{a ^{2}+b ^{2}}\]
\[\tan \theta =\frac{ b }{ a }\]
Now since these come in the form of "a+bi", whatever is not paired with i will be your "a" and what is with i is your "b". So it looks like you have:
\[a = 1 + \sqrt{3}\]
\[b = 1 \]
So using that as your a and b and using the conversion above, you think you could get r?

Answer 3

this may seem like really basic math but how do i square 1 + sqrt(3) ?

Answer 4

\[\sqrt{(1)^{2}+\sqrt{3}^{2}} = \sqrt{1+3}\]

Answer 5

Hmm....wait, nvm.

Answer 6

I did that without thinking, haha.

Answer 7

Well, we can either turn it into a decimal so we can combine it or we have to do this:
\[r=\sqrt{(1+\sqrt{3})^{2}+1}\] And then its up to you what yoiu want to do with that. But that is sqrt(a^2 + b^2). Do you want to simplify it, make it a decimal, or what? It is kind of weird since you have a problem like this, though xD

Answer 8

i just made it a decimal. I used my calculator and i got 7.464101615...

Answer 9

for (1+sqrt(3)) ^2

Answer 10

Ah. Alright, I'll do the whole a^2 + b^2 calculator work.

Answer 11

So 2.91

Answer 12

Okay so I'm getting r to equal=2.9 right?

Answer 13

oh yeah haha we both got it!

Answer 14

Are you sure theres a plus between sqrt 3 and i?

Answer 15

oh no there is not! haha sorry my fault

Answer 16

Okay, no wonder x_x Makes this much easier. Okay, so then our r is this:
\[\sqrt{(1)^{2}+(\sqrt{3})^{2}}\]

Answer 17

Which I actually posted above, lol.

Answer 18

Yes! so r=2

Answer 19

Yeah xD Much more sense. R = 2 and r is the modulus. So now we need theta. We do that using this conversion:
\[\tan \theta = \frac{ b }{ a }\]
Think you might know how to get that?

Answer 20

yes theta = sqrt(3)?

Answer 21

Tan(theta) = sqrt(3). You only have tangent, you need the actual theta xD

Answer 22

oh that means theta=60 degress!

Answer 23

Sounds good to me, lol. I guess you need it in degrees. So what if it were 1 - sqrt(3)i. Then what would theta be? (just to make sure we get the idea)

Answer 24

wouldn't it be the same thing since -sqrt3 * -sqrt3 is still 3?

Answer 25

Thats if we find r, though, not theta. Theta would definitely be different.

Answer 26

hmm I'm not exactly sure.. 300 degrees?

Answer 27

Yep xD The reason I ask is because you almost always havetwo options for theta. tan(theta) = sqrt(3) has two answers, but you need to know which is the correct one. The way you know is to look at your original rctangular problem. With us, we need to go right 1 and then up sqrt(3) units. In the example I gave you, itd be right 1 but down sqrt(3) units. So I just wanted to bring that to your attention :P

Answer 28

okay that's good to know! Do you know how to do part 3 or 4 of this question?

Answer 29

Well, 3 and 4 are almost the same problem. But okay, so this is what we should have so far for our polar form:
\[2(\cos60 + isin60)\]Now in order to find roots, we have to use this formula:
\[\sqrt[n]{r}(\cos(\frac{ \theta }{ n }+\frac{ 2k \pi }{ n })+isin(\frac{ \theta }{ n }+\frac{ 2k \pi }{ n }))\]
Where n is theroot we need. So basically we take the nth root of r and then divide each angle by whatever root what we want. After we do that, we have our first root. To get the other roots, you use the + 2kpi/n part. So if we need a 5th root, think you can find the first of the 5 5th roots?

Answer 30

Yikes this math is way beyond me... I don't really know :(

Answer 31

Well, we have
\[2(\cos60 + isin60)\]
So just take the 5th root of r and divide the angles by 5.

Answer 32

so since r is 2, we take the 5th root of 2? which is apprx 1.15 ?

Answer 33

Well, you usually don't put it into decimals unless it asks you to. Decimals we usually try to avoid if we can xD I would just leave it as:
\[\sqrt[5]{2}\] or something like that xD

Answer 34

okay :) are you supposed to divide 5th root of 2 by 4 to get part 3?

Answer 35

Oops, we needed 4th roots, lol. My bad x_x

Answer 36

\[\sqrt[4]{2}\]

Answer 37

So thats the radius of our 4th roots. Now divide the angles by 4.

Answer 38

which angles? :o

Answer 39

Did you never get or see the polar form we came up with? :P

Answer 40

oops sorry is it cos60 and sin60?

Answer 41

Yeah xD So divide those by 4.

Answer 42

so 1/8 and sqrt(3)/8

Answer 43

If you turn them into that, then yes, lol. Can you put all of that in standard form before we continue?

Answer 44

1/8 + sqrt(3)/8 i

Answer 45

No r? xD

Answer 46

oops im getting really mixed up.. i dont know where the r goes.

Answer 47

You just dont remember the standard form is all :P
\[r(\cos \theta + isin \theta)\]

Answer 48

okay 2(cos(60) + i sin(60) = 2(1/8 + sqrt(3)/8 i)

Answer 49

Sorry to keep making corrections here and there, but we have to keep it in sin and cos, we cant say cos60 = 1/2 or anything, we'll getthe wrong answers. You just gotta get into the habit of not making everything decimals or simplifying things unless we need to xD Sorry @_@

Answer 50

oh its okay! so what are the 4 angles?

Answer 51

Well, whats 60/4? :P

Answer 52

15

Answer 53

Right. So this is the first answer.
\[\sqrt[4]{2}(\cos15+isin15)\]
Now you gotta now how we get here, though. We've spenta lot of time, somake sure you know what we did to get here o.o

Answer 54

yeah i think i understand it now! would you just add another 15 degrees to get the 2nd angle?

Answer 55

Nope. Remember thewhole 2kpi/n thing?

Answer 56

oh right so add 360 degrees?

Answer 57

uh, no o.o you add:
\[\frac{ 2k \pi }{ n }\] where n is the root you want. So sincen is 4, we would add:
\[\frac{ k \pi }{ 2 }\] where k is 0, 1, 2, 3.
k = 0 was our first 4th root that we just typed. Adding that when k = 1 is the 2nd, adding that when k = 2 is the 3rd and k = 3 is the 4th. Although Im sure ill have to explain and show more than that, lol.

Answer 58

hmm okay thank you for all your help, i really really appreciate it! :D

Answer 59

Lol, you think you got it?

Answer 60

no but its really late and i dint think im gonna get it but thanks again for all your efforts! i understand it better than i did!

Answer 61

*don't

Answer 62

Well we'll get it later then, lol.

Answer 63

yeah thank you for trying! :)

Answer 64

Night, lol.