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Complex quadratic equation

  1. Feb 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Hello!

    z^2 - (7+2i)z + (12+8i) = 0

    3. The attempt at a solution

    I used the quadratic equation and got:

    z = (7+2i)/2 +- sqrt( -3/4 - i)

    r = 5/4, and the argument is Theta = arctan(-1/-3/4) = 0.927 925 - Pi (since x < 0 and y < 0, right?)

    Then, I know that z = e^theta*i, or z =r * (cos(theta) + i sin(theta)). Hence

    sqrt(z) = sqrt(r)sqrt(cos(theta) + i sin(theta))

    ...

    But then, I'm stucked...

    Any suggestion will be very appreciated.
     
  2. jcsd
  3. Feb 22, 2012 #2

    eumyang

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    Do you know De'Moivre's formula? One of its applications is finding the nth root of a complex number.
     
  4. Feb 22, 2012 #3
    sqrt(z) = sqrt(r){cos[(theta + 2Pi)/2] + i sin[(theta+ 2Pi)/2]}

    and got : -0.894 - 0.447i

    But this answer is wrong.

    Apparently in the argument Theta = arctan(-1/-3/4) = 0.927 925 - Pi

    I should be adding Pi, instead of substracting it, but I don't understand why. I read that if x < 0 and y < 0, I should substract Pi...

    that answer should be -0.5 + i

    ...
     
  5. Feb 22, 2012 #4

    eumyang

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    I think here is the confusion. If you are trying to find a trig function of a 3rd quadrant angle, you subtract π to find its reference angle. This isn't the same thing. When you take an arctan, you get an angle between -π/2 and π/2 (but not including -π/2 or π/2), so depending on where the point is on the complex plane, you may need to add π (or 2π, if the point is in the 4th quadrant you wish your angle to be in 0 ≤ θ < 2π).
     
  6. Feb 22, 2012 #5
    Ok. Let's see if I understand it now:

    Arctan gave me this angle: 0.927 925, which is on the first quadrant. But the point on the complex plane is on the third quadrant. So I would have to add pi, so that the angle reaches the third quadrant?

    "or 2π, if the point is in the 4th quadrant you wish your angle to be in 0 ≤ θ < 2π"

    But arctan includes the IV quadrant, right? -Pi/2 is not -270 degrees?
     
  7. Feb 22, 2012 #6
    i'm affraid that I'm still having the incorrect answer:

    From

    sqrt(z) = sqrt(r){cos[(theta + 2Pi)/2] + i sin[(theta+ 2Pi)/2]}

    And using the new value of arctan, 4.0688...

    I have the feeling that the mistake is somewhere over here:

    sqrt(z) = sqrt(r){cos[(theta + 2Pi)/2] + i sin[(theta+ 2Pi)/2]}

    But I can't tell...
     
  8. Feb 22, 2012 #7

    Deveno

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    one thing that might help you is this:

    we are trying to find the square root of this complex number:

    z = -3/4 - i.

    now, z = r(cosθ + i sinθ).

    you correctly calculated r as 5/4.

    but finding "θ" is bogging you down.

    all you need to know is that:

    -3/4 - i = 5/4((-3/5) + i(-4/5))

    this tells us cosθ = -3/5, sinθ = -4/5.

    now de moivre's formula lets us calculate one square root of z as:

    √z = √r(cos(θ/2) + i sin(θ/2)).

    we don't need to find θ, then divide θ by 2, and then take the cosine and sine. knowing cosθ alone is good enough, because:

    cos(θ/2) = ±√[(1+cosθ)/2]

    and we can use the quadrants to tell us which sign to take. since θ is in the 3rd quadrant, θ/2 is in the second quadrant, so we want the negative square root.

    similarly, sin(θ/2) = ±√[(1-cosθ)/2].

    so, have fun :)
     
  9. Feb 22, 2012 #8

    eumyang

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    Technically, it should be
    [itex]\sqrt{z} = \sqrt{r}\left( \cos \left( \frac{\theta + 2k\pi}{2} \right) + i\sin \left( \frac{\theta + 2k\pi}{2} \right) \right)[/itex],
    with k = 0 and 1.
    So for k = 0:
    [itex]\sqrt{z} \approx \sqrt{5/4}\left( \cos \left( \frac{4.0689}{2} \right) + i\sin \left( \frac{4.0689i}{2} \right) \right)[/itex],

    Check to make sure that you had multiplied by the square root of 5/4 at the end. I'm not sure you did.
     
  10. Feb 22, 2012 #9
    well, I must confess I'm terrible with the calculator. Now it works.

    I also tried Deveno's method and it worked also. The only thing I actually don't grasp 100 percent is the following sentence "since θ is in the 3rd quadrant, θ/2 is in the second quadrant, so we want the negative square root." I know θ is on the III quadrant, but why θ/2 is in the second? As you can see my trignometry knowledge is very weak. I could say I didn't learnt trigonometry, until very recently...

    But I liked Deveno's method! However, I found there is even a faster and easier method!

    sqrt(-3/4 - i) = a + ib

    => -3/4 - i = a^2 - b^2 + 2abi

    The real part and imaginary, then, must be

    -3/4 = a^2 - b^2

    -i = 2abi

    => b = 1 and a = -1/2

    => -1/2 + i.
     
  11. Feb 23, 2012 #10

    eumyang

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    If θ is in the 3rd quadrant, then π ≤ θ ≤ 3π/2. This means that π/2 ≤ θ/2 ≤ 3π/4, which means that θ/2 is in the 2nd quadrant.

    Shouldn't there be another answer? You're looking for two square roots.
     
  12. Feb 23, 2012 #11

    Deveno

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    well, yes.

    the other square root should be "halfway-'round" the same circle the square root we found was on. that is, the same magnitude, but in the opposite direction.

    if we call the square root we found w, then it's clear that the other one we are looking for, is -w.

    however, since the quadratic formula has a +/- in it, it takes care of this for us:

    {±w} = {±(-w)}.
     
  13. Feb 23, 2012 #12

    eumyang

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    I was asking the OP, actually. I wanted to make sure he/she knew that ultimately there were two solutions for z.
     
  14. Feb 24, 2012 #13

    Deveno

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    interestingly enough, one of the roots is real.
     
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