Expert: Paul Klarreich - 10/19/2010

Question
QUESTION: Q: Find all solutions to the equation (z^3+1)*(z^2+z+1)=0, giving all roots in polar form(that is, in the form r cis theta )with the added restriction -pi<thetha<pi

I could use completing squares to solve (z^2+z+1) but don't know how to solve (z^3+1). I was thinking:

z^3= -1(i)
z^3=(i^2)(i)
z^3=i^3
z=i

is that right? Thanks heaps =)


ANSWER: Questioner:   Caleb
Country:  Australia
Category:  Calculus
Private:  No
 
Subject:  complex numbers
Question:  Q: Find all solutions to the equation (z^3+1)*(z^2+z+1)=0, giving all roots in polar form(that is, in the form r cis theta )with the added restriction -pi<thetha<pi

I could use completing squares to solve (z^2+z+1) but don't know how to solve (z^3+1). I was thinking:

z^3= -1(i)  <<< looks wrong to me.
z^3=(i^2)(i)
z^3=i^3
z=i

is that right? Thanks heaps =)
.........................................

Have you studied deMoivre's Theorem?  Assuming you have, then

Part I:  z^3 = - 1

is solved this way:

r^3 cis 3t = 1 cis 3pi/2

and you can easily do the rest, I think.  (If not, let me know.)

..........................

Part II: Now about  z^2 + z + 1, I wouldn't use CTS for that.  Instead, note that:

(z^2 + z + 1)(z - 1) = z^3 - 1

The solutions of  z^3 = 1  are the same as the solutions of z^2 + z + 1 = 0, except for the  z = 1 solution.

So do this the same way as Part I, just exclude  z = 1 and keep the other two solutions, giving you five in all.


---------- FOLLOW-UP ----------

QUESTION: Oh i got it. Thanks a lot.
Urm, just one last question.
May i know what does this question wants?

Re((7+9i)z-3)<0
Does it mean it only wants real numbers?
I substituted z=x+yi into it as you taught me previously and i got (4+x)+i(9+y)<0.
Since i am assuming it only wants real number so is this the answer? (4+x)<0 ?
Thanks


Answer
Questioner:   Caleb
Country:  Australia
Category:  Calculus
Private:  No
 
Subject:  DeMoivre's Theorem
---------- FOLLOW-UP ----------

QUESTION: Oh i got it. Thanks a lot.
Urm, just one last question.
May i know what does this question wants?

Re((7+9i)z-3)<0
Does it mean it only wants real numbers?
I substituted z=x+yi into it as you taught me previously and i

got (4+x)+i(9+y)<0.
Since i am assuming it only wants real number so is this the

answer? (4+x)<0 ?
Thanks
......................
I think you messed it up:

(7 + 9i)z - 3  =

(7 + 9i)(x + iy) - 3 =

7x - 9y - 3 + (7y + 9x)i

Next:  Re(of that ) < 0 means

7x - 9y - 3  < 0

This is a region of the plane.  You can graph it (see attached)