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Polar coordinate

  1. Jan 12, 2006 #1
    x^2 + 2x + 5 = 0.Find the root of this eqn.can use polar system.

    is the answer=(-1+2i) @ (-1-2i)??
    pls help.......
    thanx....
     
  2. jcsd
  3. Jan 12, 2006 #2

    StatusX

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    That answer is in cartesian form. A polar answer looks like r e. Polar coordinates wouldn't really be appropriate to do this problem, although you could easily put the final answer from cartesian form, z=a+bi, into polar coordinates using r2=a2+b2, Θ=tan-1(b/a).
     
    Last edited: Jan 12, 2006
  4. Jan 12, 2006 #3
    can youpls show me the stpes??plsssss
     
  5. Jan 12, 2006 #4

    StatusX

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    I won't do it for you, but I'll answer any specific questions you have. But if you are going to use cartesion coordinates, just use the quadratic formula.
     
  6. Jan 12, 2006 #5
    i have tried it.so is it the ans is 5^1/2 e(j0.6476pai)??
     
  7. Jan 12, 2006 #6

    HallsofIvy

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    Please copy the problem exactly as it is given to you. In your first post you said
    "x^2 + 2x + 5 = 0.Find the root of this eqn.can use polar system.

    is the answer=(-1+2i) @ (-1-2i)??"

    "can use polar system" doesn't mean you have to! It's easy to solve the equation by completing the square. Yes, the solutions are
    x= -1+ 2i and x= -1- 2i. Are you required to write the answers in polar form?
     
  8. Jan 13, 2006 #7
    AFAIK, there's no polar equivalent to adding numbers, so If anyone who knows how to solve it using polar system, please let us know.

    Apart from that, it most prolly is to find the roots using quadratic formula in cartesian form ( x+iy ) and convert it to polar form

    r(e)^iD, where r=sqrt(x^2+y^2) and angle D=arctan(y/x)
     
  9. Jan 13, 2006 #8
    ya,require.....
     
  10. Jan 13, 2006 #9
    then convert using the expressions I gave in the last post.
     
  11. Jan 14, 2006 #10

    HallsofIvy

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    Polar form of the number a+ bi is either [itex]r(cos\theta+ i sin\theta)[/itex] or [itex]r e^{i\theta}[/itex] (since [itex]e^{i \theta}= cos\theta+ i sin\theta[/itex] they are equivalent) where r is |a+ bi| and [itex]\theta[/itex] is the "argument" or angle the line through (0,0) and (a,b) makes with the positive real axis. For a+ bi, [itex]r= \sqrt{a^2+ b^2}[/itex] and [itex]\theta= arctan(\frac{b}{a})[/itex] as long as a is not 0. If a is 0 and b is positive, then [itex]\theta= \frac{\pi}{2}[/itex]. If a is 0 and b is negative, then [itex]\theta= -\frac{\pi}{2}[/itex]. The number 0 (0+ 0i) cannot be written in "polar form".

    If you were given a problem requiring the answer in polar form, surely you were already taught all of that?
     
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