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frx Group Title

Solve the complex equation: \[(2+i)z ^{2}+(8-11i)z-5-25i=0\] I started to isolate the first term and got: \[z ^{2}+\frac{ 8-11i }{ 2+i }z + \frac{ -5-25i }{ 2+1 } = 0\] \[z ^{2}+(1-6i)z-(7+9i)=0\] How should i continue from here? I can't use the quadratic formula, right? So i guess i need to factorize, but how?

  • one year ago
  • one year ago

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  1. Yahoo! Group Title
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    u can use quad formula here

    • one year ago
  2. frx Group Title
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    What happens then when i take the square root of -7/4+6i?

    • one year ago
  3. frx Group Title
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    \[\sqrt{-\frac{ 7 }{4 }+6i}\]

    • one year ago
  4. Yahoo! Group Title
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    \[Z = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]

    • one year ago
  5. Yahoo! Group Title
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    \[i^2 = -1\]

    • one year ago
  6. frx Group Title
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    \[\frac{ \-1+6i \pm \sqrt{(1-6i)^{2}+4(7*9i)} }{ 4}\]

    • one year ago
  7. frx Group Title
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    But 1-6i squared is 1-12i-36 so i still get the root of i

    • one year ago
  8. hartnn Group Title
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    can't u use calculator ?

    • one year ago
  9. frx Group Title
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    We aren't allowed to use calculators at all at my university

    • one year ago
  10. hartnn Group Title
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    then its not easy to solve this: even with completing the square method, u need square root of complex number.

    • one year ago
  11. hartnn Group Title
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    can u convert complex number to polar form manually ?

    • one year ago
  12. frx Group Title
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    Yepp, maybe it's easier to solve in Euler form?

    • one year ago
  13. hartnn Group Title
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    not tried, ever

    • one year ago
  14. frx Group Title
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    Ok, I'll try it out it might just work :)

    • one year ago
  15. hartnn Group Title
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    u know the answer ? if not wanna know ?

    • one year ago
  16. frx Group Title
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    I know it, z=1+5i and z=-2+i

    • one year ago
  17. frx Group Title
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    It got pretty wierd using the Euler formula so it's probobly not useble. But the Wolframalpha solutions factorize the hole thing into \[(z-1-5i)(z+(2-i))=0\] But I don't see how to come up with that without any computer..

    • one year ago
  18. hartnn Group Title
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    yeah, even i did see wolf, all 3 methods....each required sqrt of (-7/4+6i)

    • one year ago
  19. frx Group Title
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    By completing the squares it can also be written as\[(z+\frac{ 1-6i }{ 2 })^{2}-(\frac{ 1-6i }{2 })^{2} - (7+9i) = 0\]

    • one year ago
  20. phi Group Title
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    one way to find the square root of the discriminant is use the idea (a+bi)^2 = sqrt( -117 + 44i) (Here I am using a, b, and c from the original problem) this means a^2-b^2 +2abi = -117 + 44i you can solve for a and b using a^2-b^2 = real part = -117 2ab = imag part = 44

    • one year ago
  21. frx Group Title
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    Ohh, that's clever! Will try it out!

    • one year ago
  22. phi Group Title
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    trust me, I did not figure this idea out first.

    • one year ago
  23. frx Group Title
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    I've seen something like it before in my book so I get the idea :) But is this the general way to solve these kind of problems?

    • one year ago
  24. phi Group Title
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    that idea is the basis for formulas for finding the root of a complex number in rect coords you could also change to polar. But that really needs a calculator to do the atans And in general, you are not going to get nice numbers no matter what approach (i.e. numbers you can handle without a calculator). This problem was chosen to be doable by hand (with a lot of work)

    • one year ago
  25. frx Group Title
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    I really don't get why we are not allowed to use calculators but i guess it's good for the brain to do there problems without it. Is quite hard to convert into polarform to since the argument gets really nasty. Anyway, thank you alot for your help will have to practice this for som time i guess ;)

    • one year ago
  26. frx Group Title
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    By the way are you guys at MIT allowed to use calculators to there kind of problems?

    • one year ago
  27. phi Group Title
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    I graduated a while ago. But in general I would guess yes. Lots of times tests were open book (not that it helped!)

    • one year ago
  28. frx Group Title
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    Open Book seems at least more genuine, it reflects the worklife better, than our memorization tests.

    • one year ago
  29. phi Group Title
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    your discriminant simplifies to -7+24i the square root is reasonably straight forward

    • one year ago
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