Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complex Number - Locus

  1. Sep 4, 2010 #1
    Hey i have a question:

    Q. One root of the cubic equation is z^3 + az + 10 = 0 is 1 + 2i.
    (i). Find the value of the real constant a.
    (ii). Show all the roots of the equation on an Argand Diagram.
    (iii). Show that all three roots satisfy the equation |6z - 1| = 13, and show the locus represented by this equation on your diagram.

    I did (i) and (ii) second part easily. The only problem I am facing is that of locus. I mean if the last part was like |z - 1| = 13, I know that the locus is a circle of radius 13 with center (1,0) but here it is different. I mean does 6z affect the locus. If it does then what should be the locus and if it doesnot what is the reason behind it.
     
  2. jcsd
  3. Sep 4, 2010 #2

    mathman

    User Avatar
    Science Advisor

    |6z-1|=13 is the same as |z-1/6|= 13/6, so you have a center at (1/6,0) and a radius of 13/6.
     
  4. Sep 4, 2010 #3
    Are you sure about this? I did it this way but my teacher marked it as wrong saying it should still be at center (1,0) and radius 13.
     
  5. Sep 4, 2010 #4

    Mentallic

    User Avatar
    Homework Helper

    And I bet your teacher never gave you a reason why it should still be that way.

    Ask him/her where the error is in making the following steps:

    [tex]|6z-1|=13[/tex]

    [tex]|6(z-1/6)|=13[/tex]

    [tex]|6||z-1/6|=13[/tex]

    [tex]6|z-1/6|=13[/tex]

    [tex]|z-1/6|=13/6[/tex]
     
  6. Sep 4, 2010 #5

    Mentallic

    User Avatar
    Homework Helper

    By the way, for it to be centre (1,0) and radius 13 it has to simply be |z-1|=13
    Why would we just ignore the change from z to 6z?
     
  7. Sep 5, 2010 #6
    Your point is absolutely valid if it had to be center (1,0) and radius 13 then why change z to 6z. I will ask him and you are right he never told us why it was that way. Thanks for your help. It is really appreciated!
     
  8. Sep 5, 2010 #7

    Mentallic

    User Avatar
    Homework Helper

    It's just that I was able to empathize for your position in this. Two years ago when I was in school we were short on books in our complex numbers class so I didn't have one. So instead I would get help solely from the teacher. Naturally, anything she told me I took as completely right since I haven't ever doubted a teacher in maths before. Maths only has the right way and the wrong way, and when your teacher is sure of herself, what's there to doubt, right?

    So she told me when I was studying locus that when graphing arguments, such as [itex]arg(z)=\pi/4[/itex] All I have to do is draw the line y=x and I'm done. She never told me about the restrictions or she never knew, probably the latter... So then when our teacher was replaced when we started on another topic and had a test on complex numbers again, obviously I get it wrong. It wasn't the few marks I lost that got to me, it's that teachers tend to just think they're right about everything so whenever they teach you something, they say it with utmost confidence as if there is no doubt about it.

    And then when our second teacher didn't even know how to teach these two topics in our class, I came to Physics Forums for help :biggrin: At least he was willing to admit it rather than teaching us everything the wrong way.
     
  9. Sep 6, 2010 #8
    hey listen i got it. I think it was wrong because the question says that show that the roots satisfy the equation so you need to show that it satisfies the equation. Although the equation |z - 1/6| = 13/6 also satisfies it but the question specifically asks us to do so with the equation it has given n when v check the roots v see that it satisfies so the locus should be of radius 13 at center (1,0)
    Btw its roots are 1 + 2i,1 - 2i and 2. The value of a is 1
     
  10. Sep 6, 2010 #9

    Mentallic

    User Avatar
    Homework Helper

    For z=1+2i, 6z-1=6(1+2i)-1=5+12i

    [itex]|6z-1|=\sqrt{5^2+12^2}=13[/itex] so it's right.

    [itex]|z-1/6|=13/6[/itex] also satisfies because they are the same, we have manipulated them using correct algebraic techniques. Cutting off the 6 in front of the z is not a correct algebraic technique...

    |z-1|=13 does not satisfy the roots, thus the roots do not lie on a circle centre 1 radius 13.
     
  11. Sep 6, 2010 #10

    Mentallic

    User Avatar
    Homework Helper

    By the way, the real root is -2, not 2.
     
  12. Sep 8, 2010 #11
    yeah you r right the root is -2 i missed it while typing. n thanx now i got a perfect reason to debate.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook