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Imaginary numbers

  1. Apr 16, 2003 #1
    Could someone PULEEZ explain how to work the following equation:

    3-7i/2+3i

    For the life of me I cannot sqeeze this into my brain!

    Thank you in advance.
     
  2. jcsd
  3. Apr 16, 2003 #2
    Have you tried multiplying both the Numerator and Denominator by the complex conjugant of the Denominator?
    …Don’t forget to use the FOIL method.
     
  4. Apr 16, 2003 #3

    enigma

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    Huh? That doesn't look like a multiplication problem to me, Boulder.

    Echo, when adding and subtracting complex terms (imaginary and real), you just keep the like terms together.

    In other words, you add the imaginary parts, and then you add the real parts.

    For example:

    3+8+5i+7i/4=

    (3+8)+(5+7/4)i=

    11+(27/4)i
     
  5. Apr 16, 2003 #4
    We had conversed through PM and Echo 6 Sierra didn't make me aware of this. We treated the problem as;

    (3-7i)/(2+3i)

    Perhaps Echo 6 Sierra can clarify, as I assumed the parenthesis had been mistakenly omitted. If not, my mistake then.
     
    Last edited by a moderator: Apr 16, 2003
  6. Apr 16, 2003 #5

    enigma

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    Ah hah.

    Gotcha. Parenthesis. Yes, that would be division then, wouldn't it?

    :wink:
     
  7. Apr 16, 2003 #6
    Yes, I apologize. I mistakenly omitted them. Also, another apology is in order. I mistakenly posted here instead of the Homework section. Thank you both for your input.
     
  8. Apr 16, 2003 #7

    dav2008

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    Hm...Did you solve it yet?


    But yea, conjugate of the denomitator is how u do it...

    OR

    You can change it to polar ( i think thats what its called) And then u have
    r*cis([0])
    r2*cis([0]2)
    and u just divide the radii and subtract the angles...Not sure tho..i know for multiplication u multiply the radii and add the angles so i would assume u do the opposite for division
     
  9. Apr 17, 2003 #8
    Problem solved. Thank you to everyone for your input.
     
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