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Surds in polar form of imaginary number

  1. Feb 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the polar form for zw by first putting z and w into polar form.
    z=2√3-2i, w= -1+i

    2. Relevant equations
    Tan-1(-√3/3)= 5∏/6

    3. The attempt at a solution
    r= √[(2√3)2+(-2)2]=4
    tanθ= -2/(2√3)=-1/√3=-√3/3=> acording to above... tan-1(-√3/3)= 5∏/6
    so, in polar form z should be 4[cos(5∏/6)+isin(5∏/6)]...
    However, in the markscheme as well as in another source i have (thus me deducing there was no coincidence) they've put 4[cos(-∏/6)+isin(-∏/6)].
    I understand how tan(5∏/6) and tan(-∏/6) both equal(-√3/3)... though that's not the case for when it comes to cos(5∏/6) and cos(-∏/6).
    So i would appreciate it if someone explained to me where i have gone wrong... and i think the question is pretty clear but if you think i haven't explained myself correctly go forth and tell me!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 22, 2013 #2

    CAF123

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    Sketch ##z = 2 \sqrt{3} -2i## in the Argand plane. What quadrant does it lie?
     
  4. Feb 22, 2013 #3

    HallsofIvy

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    Tangent is not a "one-to-one" function and so arctangent is not "well-defined". In fact, you calculator should show you that [itex]tan(5\pi/6)= tan(-\pi/6)[/itex]. Which quadrant is [itex](2\sqrt{3}, -2)[/itex] in? Where are [itex]5\pi/6[/itex] and [itex]-\pi/6[/itex]?
     
  5. Feb 22, 2013 #4
    I'm new to imaginary numbers so bare w me here...
    z=2√3−2i in the Argand plane lies in the 4th quadrant...?

    5π/6 lies in the 2nd Q. and −π/6 doesn't lie on any Q? π/6 lies in the first quadrant. What's the next step?
     
  6. Feb 22, 2013 #5

    CAF123

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    So if your complex number lies in the 4th quadrant, what is the appropriate argument?

    Do you see how the other argument arises?
     
  7. Feb 22, 2013 #6
    hun? im dont understand what you're saying.... If i look at the 4th quadrant of the trigonometric circle then wouldn't it be 11pi/6? which i guess is the reflection of the pi/6 on the x-axis?? as i've told you, im not sure how to work myself around this yet
     
  8. Feb 22, 2013 #7

    HallsofIvy

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    The trig functions, and so the polar representation of complex numbers, is "modulo [itex]2\pi[/itex]". That is [itex]-\pi/6[/itex] is exactly the same as [itex]2\pi- \pi/6= 11\pi/6[/itex].
     
  9. Feb 22, 2013 #8
    that makes sense. so i would guess that adding 11pi/6 or -pi/6 to something (pheta1 + pheta2) if i were to be multiplying to complex numbers in polar form then i would get the same result?
     
  10. Feb 22, 2013 #9
    4[cos(-∏/6)+isin(-∏/6)]*2[cos(3∏/4)+isin(3∏/4)] = 4[cos(11∏/6)+isin(11∏/6)]*2[cos(3∏/4)+isin(3∏/4)] ?
     
  11. Feb 22, 2013 #10
    could someone confirm this for me? thanks
     
  12. Feb 22, 2013 #11

    CAF123

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    Yes, the two angles represent the same point, but generally we choose the argument such that ##arg(z) \in [0,\pi],##or##[0,-\pi]##

    This is why the book gives the argument -pi/6 in the solution.

    The reason you got 5pi/6, which also satisfies arctan(##-1/\sqrt{3})## is because there we deal with ##z = -2\sqrt{3} + 2i##. So blindly computing arctan(..) without thinking where the complex number lies causes problems.
     
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