Roots of a Complex Number

Au101 · 2011-06-26 04:11:04

Hi guys,

I have a bit of a problem with the following question:

My answers to part a. are as follows:

Which I believe to be correct. The answer book actually gives the answer as:

However, I do not think that this is right. Even if this is right, it doesn't help me with part c. The answer which is given is that the centre is (0,0) and the radius 1. 1, of course, is the magnitude of both:

Which makes sense, however, how can these points form a circle when the third root is at -2. Surely the radius must be 2 - I simply don't understand

Thanks

Last edited by Au101 (2011-06-26 04:12:09)

anonimnystefy · 2011-06-26 04:27:25

hi Au101

for (a) probably a typing error

as for c i think that the centre is in (-1,0) and the radius is 1.

gAr · 2011-06-26 04:35:10

Hi Au101,

I agree with anonimnystefy.

Your answer's correct.
Center's at (-1,0) with radius 1.

Au101 · 2011-06-26 04:36:56

Hello, there, anonimnystefy - I agree with you on part a. but it's nice to have my answered confirmed. As for your answer to part (c) - that makes an awful lot of sense, I think that you're absolutely right. I can definitely justify that answer, but I'm not sure how I would go about deriving it. Could you advise me on how to set-out an answer?

Edited to add: Hi gAr - sorry, I was replying when you replied. Thanks very much . I don't suppose that you could help with my second question as well?

Thanks a lot:)

Last edited by Au101 (2011-06-26 04:37:55)

anonimnystefy · 2011-06-26 04:48:10

hi Au101

i did it with the hep of the roots of unity.then i took a symetric picture of the circle with the centre being the vertical line that goes through the point (1/2,0) and showed that it's the circle that we need.

Bob · 2011-06-26 05:05:49

hi Au101,

Your answers are definitely correct. The equation you are solving has real coefficients so it has to have one real root and a pair of complex conjugates.

The 'answer book' answers aren't conjugates. We'll let them off with a typo shall we?

Part b then nudges you towards what follows:

The complex pair are equi-distant from (-1,0) and (0,0) so either is a possible contender for a centre going through both. The third point forces us to accept (-1,0).

You are told that they lie on a circle so you don't have to prove that from scatch; it is acceptable to find a circle with the right properties and that's it done.

Bob

Last edited by Bob (2011-06-26 05:09:19)

Au101 · 2011-06-26 05:18:01

Perfect - thank you everyone

gAr · 2011-06-26 05:19:21

Hi Au101,

a)
The way I solved was using de Moivre's theorem:

And I agree with what Bob said.

Math Is Fun Forum

#1 2011-06-26 04:11:04

Roots of a Complex Number

#2 2011-06-26 04:27:25

Re: Roots of a Complex Number

#3 2011-06-26 04:35:10

Re: Roots of a Complex Number

#4 2011-06-26 04:36:56

Re: Roots of a Complex Number

#5 2011-06-26 04:48:10

Re: Roots of a Complex Number

#6 2011-06-26 05:05:49

Re: Roots of a Complex Number

#7 2011-06-26 05:18:01

Re: Roots of a Complex Number

#8 2011-06-26 05:19:21

Re: Roots of a Complex Number

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