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mathslover

  • 2 years ago

Show that every complex number is a root of some quadratic equation.

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  1. waterineyes
    • 2 years ago
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    I forgot to say : Congratulations for \(\large \color{green}{\frak{green..}}\) @mathslover

  2. mathslover
    • 2 years ago
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    Thanks waterineyes.

  3. klimenkov
    • 2 years ago
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    \((x-(a+bi))(x-(a-bi))=x^2-2ax+(a^2+b^2)\) So, we get that this quadratic equation has a root \(x=a+bi\). So as \(a\) and \(b\) are any real numbers we prove it for any complex number.

  4. mathslover
    • 2 years ago
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    Hmn wait let me check it .. :)

  5. Mikael
    • 2 years ago
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    What' remarkable in @klimenkov is that his coefficients are REAL

  6. mathslover
    • 2 years ago
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    Oh k so there is a mistake. That is : Where is quadratic equation @klimenkov ? it must have been : \[\large{(X-(a+ib))(X-(a+ib))=\color{blue}{0}}\]

  7. mathslover
    • 2 years ago
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    Ok I meant (a-ib) in one of them .. :)

  8. UsukiDoll
    • 2 years ago
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    ummm what subject is this?

  9. klimenkov
    • 2 years ago
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    1) Equation \(x^2-2ax+(a^2+b^2)\) is QUADRATIC. 2) Equation \(x^2-2ax+(a^2+b^2)\) has a root \(x=a+bi\). 3) \(a+bi\) is any complex numer for \(a\in \mathbb R, b \in \mathbb R\).

  10. klimenkov
    • 2 years ago
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    Try to multiply carefully, there is no mistake.

  11. mathslover
    • 2 years ago
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    @klimenkov I was saying that, you didn't include 0 in RHS, why so?

  12. klimenkov
    • 2 years ago
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    Ah.. I'm really sorry. You are absolutely right. It was not an equation. It was just a polynomial.

  13. mathslover
    • 2 years ago
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    Oh no problem. That was a minor mistake ... No problem.. Thanks though.

  14. waterineyes
    • 2 years ago
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    @UsukiDoll this is Mathematics...

  15. klimenkov
    • 2 years ago
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    I like questions from people who have SmartScore > 75. Ask something interesting again!

  16. mathslover
    • 2 years ago
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    OK, if I will get a doubt, then I will like to ask you.

  17. AravindG
    • 2 years ago
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    am i late?

  18. klimenkov
    • 2 years ago
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    But if you have no need in real coefficients, you can just make an equation \((x-(a+bi))^2=0\)

  19. AravindG
    • 2 years ago
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    @mathslover i would just observe the graph and stat this statement is true

  20. AravindG
    • 2 years ago
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    and would keep in mind that evry real number a can be written as a complex number a+0i

  21. UsukiDoll
    • 2 years ago
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    @wwaterineyes besides Mathematics. I meant what level of Mathematics is this because this certainly doesn't look like calculus or any lower level math question.

  22. waterineyes
    • 2 years ago
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    If you are asking for chapter then it is \(\mathsf{Complex \; \; Numbers...}\)

  23. UsukiDoll
    • 2 years ago
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    o_O

  24. waterineyes
    • 2 years ago
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    Ha ha ha ha...

  25. waterineyes
    • 2 years ago
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    And I was just kidding...

  26. klimenkov
    • 2 years ago
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    If you are asking for chapter then it is \(\text{Algebra}\).

  27. UsukiDoll
    • 2 years ago
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    oh my O_O

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