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frx

  • 2 years ago

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?

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  1. Yahoo!
    • 2 years ago
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    u can use quad formula here

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

  3. frx
    • 2 years ago
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    \[\sqrt{-\frac{ 7 }{4 }+6i}\]

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

  5. Yahoo!
    • 2 years ago
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    \[i^2 = -1\]

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

  7. frx
    • 2 years ago
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    But 1-6i squared is 1-12i-36 so i still get the root of i

  8. hartnn
    • 2 years ago
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    can't u use calculator ?

  9. frx
    • 2 years ago
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    We aren't allowed to use calculators at all at my university

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

  11. hartnn
    • 2 years ago
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    can u convert complex number to polar form manually ?

  12. frx
    • 2 years ago
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    Yepp, maybe it's easier to solve in Euler form?

  13. hartnn
    • 2 years ago
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    not tried, ever

  14. frx
    • 2 years ago
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    Ok, I'll try it out it might just work :)

  15. hartnn
    • 2 years ago
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    u know the answer ? if not wanna know ?

  16. frx
    • 2 years ago
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    I know it, z=1+5i and z=-2+i

  17. frx
    • 2 years ago
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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..

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

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

  20. phi
    • 2 years ago
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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

  21. frx
    • 2 years ago
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    Ohh, that's clever! Will try it out!

  22. phi
    • 2 years ago
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    trust me, I did not figure this idea out first.

  23. frx
    • 2 years ago
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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?

  24. phi
    • 2 years ago
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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)

  25. frx
    • 2 years ago
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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 ;)

  26. frx
    • 2 years ago
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    By the way are you guys at MIT allowed to use calculators to there kind of problems?

  27. phi
    • 2 years ago
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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!)

  28. frx
    • 2 years ago
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    Open Book seems at least more genuine, it reflects the worklife better, than our memorization tests.

  29. phi
    • 2 years ago
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    your discriminant simplifies to -7+24i the square root is reasonably straight forward

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