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Math Help - Imaginary Numbers? Very confusing?

  1. #1
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    Imaginary Numbers? Very confusing?

    In the set of real numbers, negative numbers do not have square roots. A new kind of number, called imaginary was invented so that negative numbers would have a square root. These numbers start with the number i, which equals the square root of -1, or i2 = -1.

    All imaginary numbers consist of two parts, the real part, b, and the imaginary part, i. Example:


    1. Simplify: SQRT(-5)

    Solution: Write -5 as a product of
    prime factors.

    SQRT(-1 * 5)

    Write as separate square roots.

    (SQRT(-1))(SQRT(5))

    By definition, i = SQRT(-1),
    so the final answer is
    (SQRT(5))i. (SQRT(5)
    is the real part, or b.)

    I just dont get this entire thing... how does it even work?
    Last edited by Gordon; August 4th 2010 at 08:31 AM.
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  2. #2
    Senior Member yeKciM's Avatar
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    everything with "i" is imaginary part

     a+ib

    so every number without "i" is real part , and every number with "i" is imaginary part


    if u have let's say

    z= a + b+ c + i a + i b + i c = (a+b+c) +i(a+b+c)

    so first brackets are real part of complex number and second (with "i") is imaginary part of complex number
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  3. #3
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    But where did they get SQRT(-1 * 5)? It doesn't make any sense to me though
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    Do you realize that entire college courses are given on complex numbers.
    There also may valid and yet different approaches to this question.

    Here is one. The equation x^2+1=0 has no real solutions.
    If we introduce a new a symbol i having the property that i^2=-1.
    Now we have a solution to that equation. Do you see that -i is also a solution?

    That is brief. Does it help?
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  5. #5
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Plato View Post
    Here is one. The equation x^2+1=0 has no real solutions.
    If we introduce a new a symbol i having the property that i^2=-1.
    Now we have a solution to that equation. Do you see that -i is also a solution?
    hehehe solutions are  i and -i

    but i "think" that he asked if in i \sqrt{5} , \sqrt{5} is real part or imaginary
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    I understand what your saying, but i dont understand how my question worked? you gotta be detailed with me here, and i'm going to be High School Sophmore. So i still need help.. please
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    Quote Originally Posted by Gordon View Post
    I understand what your saying, but i dont understand how my question worked? you gotta be detailed with me here, and i'm going to be High School Sophmore. So i still need help.. please
    If you did in fact understand the purpose of the symbol i then we can proceed.
    The expression \sqrt{-5} does not exist in the real numbers.
    If it did, it would solve the equation x^2+5=0. Right?
    Now (\sqrt{5}~i)^2=(\sqrt{5})^2(i)^2=(5)(-1)=-5.
    So  \sqrt{-5} can be identified as  \sqrt{5}~i a solution to x^2+5=0
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  8. #8
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    Quote Originally Posted by Plato View Post
    If you did in fact understand the purpose of the symbol i then we can proceed.
    The expression \sqrt{-5} does not exist in the real numbers.
    If it did, it would solve the equation x^2+5=0. Right?
    Now (\sqrt{5}~i)^2=(\sqrt{5})^2(i)^2=(5)(-1)=-5.
    So  \sqrt{-5} can be identified as  \sqrt{5}~i a solution to x^2+5=0
    I think i get what you are saying, do you have some practice problem so you can test me out?
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    Can you do the three operations?
    a) (2-i)+(3+2i)=?

    b) (2-i)(3+2i)=?

    c) (2+3i)^2=?
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  10. #10
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    Quote Originally Posted by Plato View Post
    Can you do the three operations?
    a) (2-i)+(3+2i)=?

    b) (2-i)(3+2i)=?

    c) (2+3i)^2=?
    A.4

    B.3

    C.12

    I have no confidents in these answers but are they right?
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  11. #11
    Senior Member eumyang's Avatar
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    OP: the rule that you're using is this:
    If b > 0, then \sqrt{-b} = i \sqrt{b}

    Anyway, I think problems like these would be better for the OP to try first:
    \sqrt{-16}

    \sqrt{-44}

    -\sqrt{-7}

    -\sqrt{-81}

    -\sqrt{-50}
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  12. #12
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    a) (2-i)+(3+2i)=5+i

    b) (2-i)(3+2i)=6+4i-3i-2i^2=8+i

    c) (2+3i)^2=4+12i+9i^2=-5+12i

    Maybe you should slow down and study this material with a live person.
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  13. #13
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    Quote Originally Posted by yeKciM View Post
    hmmmm.... don't really know meaning of "high-school rising sophomore" lol back in my hihg-school if we didn't know this "in every day or night" we couldn't get even 2 (yours D) sorry if isn't appropriate but i thought just to help
    (he can't do any complex equation without knowing this... even what "Plato" gave him to do )
    In North America a rising sophomore is @14 years old.
    He/she may have had two years of algebra, most likely one year.
    I do not know of a single secondary textbook that would use ordered pair notation for complex numbers even in advanced courses.
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  14. #14
    Senior Member eumyang's Avatar
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    Quote Originally Posted by Plato View Post
    In North America a rising sophomore is @14 years old.
    Actually, more often than not it's 15, at least in the US. It depends, because different school systems will have different rules on the cut off birthdate for registration into kindergarten.
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  15. #15
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by Plato View Post
    In North America a rising sophomore is @14 years old.
    He/she may have had two years of algebra, most likely one year.
    I do not know of a single secondary textbook that would use ordered pair notation for complex numbers even in advanced courses.
    thank you for clarification don't know what to say but here is also high-school "first grade" between 13 and 14 years old kids and when acquainted to let's say like here with complex numbers wee all have to know definitions (not just this but everything that we learned) to the word (not just to the word, we have expression "to the comma", meaning that we couldn't say definitions with "our words"). We don't use ordered pair notation for complex numbers (just for defining them, and defining operations on them), without ordered pair notation u can't explain anyone what exactly are complex numbers...
    I really don't know how someone can do that equations without knowing what is he doing exactly, that make no sense and in my mind is wrong... to teach someone something without explanation what is he doing is .. i don't know, can't think like that

    without good base of knowledge, with everything that he learn in high-school (to understand what's he learning) there is going to be huge problems if he decides to go a collage or something even more ...

    sorry if i have done some problems with this


    P.S. I edited that post there with definition of some operations on complex numbers that we needed to know to teacher even give us a math problem to solve with equations with complex numbers (basic equations)
    Last edited by yeKciM; August 4th 2010 at 10:53 AM.
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