Question

what is theta if my complex root is z=-5i?? i keep getting an error on my calculator

Answer 1

Well, you should get undefined :P

Answer 2

BUT

Answer 3

The conversion to get theta is
\[\tan \theta = \frac{ b }{ a }\]
In this case, b is -5 and a is 0. So yeah, you get:
\[\tan \theta = \frac{ -5 }{ 0 }\]
Well what that tells me is I want to know where tangent is undefined. So what angle of tangent gets me an undefined answer?

Answer 4

i know but the problem asks for a complex fifth root -_- :/

Answer 5

thank you for that explanation.. how would you solve this to get Zo?? that's all i need to solve this

Answer 6

I was going to work up to that, sorry if you didn't need any of the previous stuff. Well, if you want a root you need this:

\[\sqrt[n]{r}(\cos \frac{ \theta }{ n }+isin \frac{ \theta }{ n })\]
So you shouldn't get an error, you should just take the 5th root of r and then divide the angle by 5.

Answer 7

And if I frustrate you then just ask me to leave, I'm not out to annoy people on here.

Answer 8

but isn't r in that equation \[\sqrt{a^2+b^2}\]

Answer 9

oh sorry no you're not frustrating at all!

Answer 10

That gets you r at first, yes. But once you get r, you then take the 5th root of it.

Answer 11

I'm frustrated with my brain and the problem -__-

Answer 12

I'm not frustrated lol sorry again

Answer 13

So assuming you've made it far enough to get r and theta, we would have:
\[5(\cos \frac{ 3\pi }{ 2 }+isin \frac{ 3\pi }{ 2 })\]
Is this what you got before worrying about the roots?

Answer 14

yup. I actually got stuck when i tried to find theta.. i forgot this formula. Thank you for reminding me :)

Answer 15

Oh xD Lol, gotcha. Well now you just need to take the 5th root of r, which is 5, and then divide the angles by 5. Do you need all 5 complex roots?

Answer 16

yes, i need to answer in standard form from Zo-Z4. If you want to provide answers that's kind of you but I will only use them to check my work :) that's such a relief! thank you!!

Answer 17

Well, given the formula, what'd we get for the 1st root?

Answer 18

one quick question..the formula in my head is \[Zo=\sqrt[5]5{?}(\cos \frac{ \theta+2\pi(0) }{ n })+isin (\frac{ \theta+2\pi (0) }{ n })\] is that correct? your formula looks a bit different

Answer 19

Well I didn't throw in the next roots, but yes, that looks fine. So you would essentially divide 2pi by 5 and then add that to your current root until you gt all 5 you need. So your first root should have had 3pi/10, correct?

Answer 20

sorry one sec.. i'm rechecking my calculations :)

Answer 21

no worries.

Answer 22

ok my answer doesn't add up. I was taught to calculate theta by tan(b/a) and that gives my undefined. so i'm back to square one :/ I'm sorry, can you explain another way i can calculate theta?

Answer 23

Right, it should be undefined. But there are two angles where tangent is SUPPOSED to be undefined. Do you know those two angles?

Answer 24

pi/2 and 3pi/2 oh my.. you are a life save :O i didn't even see that!!!!!! and i have the table memorized :O

Answer 25

Lol, yeah xD Now since our rectangular point was -5, this means we are down 5 and thus at 3pi/2. Now you can continue :P

Answer 26

so because the 5 is negative.. only 3pi/2 applies? i've never seen that before *_*

Answer 27

OK! i think i have my first answer! 3pi/10? and i just add 3pi/10 to each argument after that. correct?

Answer 28

You don't add 3pi/10, you add 2pi/n. So you would add 2pi/5

Answer 29

7pi/10 is my next theta correct? before i add it to 2pi and divide by 5? i hope i'm not confusing you

Answer 30

Right, 3pi/10 is the first, 7pi/10 the 2nd, 11pi/10 the first, etc xD

Answer 31

you are seriously amazing. i wish you were my teacher :)

Answer 32

Thank you to unimaginable ends!!

Answer 33

Nah, Im not nearly as good as most people on here, lol.

Answer 34

I think I'm one of the least knowledgable to be honest, lol. People make me feel stupid all thetime x_x

Answer 35

oh no no no no. please allow yourself to receive this one complement :) you really helped me sort out a 2 hour problem!!!! I calculated Z1= cos (27/50)+iSin (27/50) in standard form. does that look correct from your work? :)

Answer 36

Where is this coming from? Just curious since we lost our r and our pi and all of that, lol.

Answer 37

because i am now adding theta to 2pi(K) all over 5. when i added Cos 7pi/10 to 2pi(K=1) all over 5 i got 27pi/50.. i knew it looked wrong

Answer 38

So you accidentally left out pi then. That and no idea where your radius went. Okay, so this is what we have:
\[\sqrt[5]{5}(\cos (\frac{ 3\pi }{ 2 }+\frac{ 2k \pi }{ 5 })+isin(\frac{ 3\pi }{ 2 }+\frac{ 2k \pi }{ 5 })\]
Sorry bout thewait. So know you just make k be 0, 1, 2, 3, 4.

Answer 39

Oops xD, should have divided the 2's by 5 as well.

Answer 40

That's ok! i can follow! i just get confused when k=1. that means i have to add 7pi/10 to 2pi and put ALL of that over 5. correct? that gives me 27pi/50

Answer 41

No, you keep constantly adding 4pi/10 each time.
So you start with 3pi/10, which is your first.
2nd = 3pi/10 + 4pi/10 = 7pi/10
3rd = 7pi/10 + 4pi/10 = 11pi/10
4th = 11pi/10 + 4pi/10 = 15pi/10
5th = 15pi/10 + 4pi/10 = 19pi/10

And those are your answers,

Answer 42

so i don't add the 2pi(k) to my new argument? sorry if i'm tiring you, i'm trying to understand this :S

Answer 43

No, you only had 2kpi/5 and you just keep constantly adding that same amount.

Answer 44

so my new answers in standard form look like:

\[z1=(\cos7\pi/10)+(isin7\pi/10)\] and

\[Z2: (\cos 11\pi/10)+(isin 11\pi/10)\]

Answer 45

correct?

Answer 46

You keep dropping your r in front xD

Answer 47

oh add that in the front mentally lol sorry! my brain is understanding stuff XD

Answer 48

Yeah, youre doing it right justcant drop the 5th rt(5) xD

Answer 49

thank you!! i understand it now :)

Answer 50

Glad to hear it ^_^

Answer 51

:))))