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Homework Help: QUICK HELP complex numbers

  1. Nov 23, 2004 #1
    Let Z1 = 3-i
    Z2=7+2i express (1/Z1)-(1/Z2) in form a+bi
    SOMEONE plzzzzzzz HELP ME!!! I dont have a clue as to how to do this :cry:
    What do I do?
    Where do I start? :cry:
  2. jcsd
  3. Nov 23, 2004 #2
    Have you ever rationalized the denominator of a fraction before? This is pretty much the same thing. For instance, when you have a fraction such as [tex]\frac{5}{\sqrt{3}+2}[/tex], you multiply the top and bottom by the conjugate, because you know a difference of squares will result in a rational number, because the square root of a rational squared is a rational. The same idea applies here: just keep in mind that [tex]i=\sqrt{-1}[/tex], and apply the same concept.

    Hope that helps! :)
  4. Nov 23, 2004 #3
    Do you know what a "complex conjugate" is?
  5. Nov 23, 2004 #4
    Yes I do it is the opposite sign well when dividing you take the denominator and divide by the conjugate I know that but in this question I dont know what to do or what order to do it in plz help me plzzz
  6. Nov 23, 2004 #5


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    nolachrymose described it pretty well.

    If you substituted i for the square root of 3,


    and multiplied by the conjugate, you'd get:

    [tex]\frac{5(\sqrt{-1}-2)}{-1 - 4} = 2 - \sqrt{-1}[/tex]

    The i is square root of negative one. If you have 3i and square it, you get -9. Other than keeping the negative signs straight, it's just like working with a square root.
  7. Nov 23, 2004 #6
    Ive never rationalized the denominator of a fraction :frown: so im a little lost I tried multiplying 1/3-i first by the conjugate and got the answer of 3+i/10 and then i did the same for z2 and got 7-2i/53 but if i subtract the two i dont get the right answer plz show me how do i divide first subtract use conjugates or what?
  8. Nov 23, 2004 #7
    As far as I can tell you rationalized the two properly. Unless you made a subtraction error, your answer should be right. What did you get as the final answer?
  9. Nov 23, 2004 #8
    There are 4 possible answers




    I dont get anything close to these when i take those two answers and subtract them. Please help me Ive been doing this one forever :mad:
  10. Nov 24, 2004 #9


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    All of their choices are wrong.

    So what did you get? Something close to

  11. Nov 24, 2004 #10


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    I think I know where they made their mistake. You can do this two ways.












    [tex]\frac{(92+(-3))+(69i+4i)}{529-(-1)}[/tex] Here's where they made their mistake


    You can break this up into two separate fractions, if you want:


    Instead of multiplying (23-i)(23+i) and getting 529+1, they got 23+1. So, the answer they most likely picked is (b)
  12. Nov 24, 2004 #11


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    Dearly Missed

    A great example of why multiple choice exams are (IMO) inferior to some other exam types which are less sensitive to designer flaws.
  13. Nov 24, 2004 #12


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    A flaw?! Or is it really a hidden 'feature'? :devil: (how come we don't have a 'shifty eyed' smilie?)

    "B's for everyone who answers all the questions right. A's for everyone who catches my error and figures out exactly where I made my mistake!"
  14. Nov 24, 2004 #13
    Well my answer was (89-33i)/530
    u are right that the question that was correct was b with the error in it. I think our answers are different because when u expanded -10(7-2i)/530 u wrote -70-20i shouldnt this be -70+2i because of the two negative signs? I will for sure point this question out to my teacher maybe she will give me an "A" lol for cathching the errors. Thanks so much I thought I was doing this totally wrong, but I guess not thanks again ur my HERO :smile:
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