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Calculating the roots of a quadratic with complex coefficien

  1. Oct 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Calculating the roots of a quadratic with complex coefficients

    2. Relevant equations
    x^2 - (5i+14)x+2(5i+12)=0

    3. The attempt at a solution

    I tried the quadratic solution but it gives too complicated solutions. I have no idea on how to do this...
     
  2. jcsd
  3. Oct 24, 2015 #2

    Student100

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    You have something in the form of $$z^2=c$$ where $$c=a+bi$$ assume that a square root of c is y+zi, where $$(y+zi)^2=a+bi$$ Do you understand where to go from here? There are multiple methods to solve these problems, some easier than others.

    Have you learned DeMoivre’s Theorem yet?
     
  4. Oct 24, 2015 #3
    No, I didn't learn it.
     
  5. Oct 24, 2015 #4
    The c^2=a+bi corresponds to what I had earlier, right ?
     
  6. Oct 24, 2015 #5

    Student100

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    Okay.

    Is the first part at least recognizable, do you know how to develop it into a formula for y and z?
     
  7. Oct 24, 2015 #6

    Student100

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    No you just had c, you're looking for $$\sqrt(c)$$
     
  8. Oct 24, 2015 #7
    Sorry, I was referring to this : z^2=18.75+25i

    ISn't this good ?

    where z=x-2.5i-7
     
  9. Oct 24, 2015 #8

    Student100

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    Looks okay.

    You're looking for a $$(y+zi)^2=18.75+25i$$ equate the real and imaginary parts.
     
    Last edited: Oct 25, 2015
  10. Oct 24, 2015 #9
    I don't understand. You want me to do this : y=18.75 and zi=25i ???
     
  11. Oct 24, 2015 #10
  12. Oct 24, 2015 #11

    Student100

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    I just made it up, it's a dummy variable.

    No, hang in there with me for a second, I want you to expand ##(y+zi)^2## first. Then equate the parts that still have an i to the imaginary part, and the parts of the equation that have no i to the real part as two equations.
     
  13. Oct 24, 2015 #12
    Ok, one step at a time : http://www4f.wolframalpha.com/Calculate/MSP/MSP64521di37fi0318f781f00005eih08d4i137gf4g?MSPStoreType=image/gif&s=22&w=132.&h=18. [Broken]

    Which gives y^2+2yzi-z^2
     
    Last edited by a moderator: May 7, 2017
  14. Oct 24, 2015 #13

    Student100

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    Excellent. Now combine like terms and tell me which are the real and complex parts and equate them to the right side respectfully.
     
    Last edited by a moderator: May 7, 2017
  15. Oct 24, 2015 #14
    y^2+2yzi-z^2

    y^2-z^2 is the real part and 2yzi is the complex part.
     
  16. Oct 24, 2015 #15

    Student100

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    So the real part is equal to your first equation, and the complex part is equal to second, why don't you go ahead and write that out.
     
  17. Oct 24, 2015 #16
    y^2-z^2=18,75

    2yzi=25i
     
  18. Oct 25, 2015 #17

    Student100

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    Yes, so a=18.75, and b =25.

    So now you have two questions, solving equation 2 for $$z=\frac{b}{2y}$$ subbing it into equation one gives you $$y^2-(\frac{b}{2y})^2=a$$ which can be rewritten as $$4y^4-4ay^2-b^2=0$$ taking the positive root $$y^2=\frac{a+\sqrt{a^2+b^2}}{2}$$ which gives you $$y=\frac{1}{\sqrt{2}}\sqrt{a+\sqrt{a^2+b^2}}$$

    That's the part of your real root, see if you can develop the complex part.
     
  19. Oct 25, 2015 #18

    PeroK

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    You shouldn't have given up on the quadratic formula so easily. With complex numbers, things tend to get algebraically messy, so you have to accept more complications than with real numbers. In any case, using the quadratic formula directly gives:

    ##x = \frac{14 + 5i \pm 5 \sqrt{3 + 4i}}{2}##

    Which is really not very complicated at all!
     
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