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    for something like

    \dfrac{3-5i}{2i}

    why not multiply by 2i/2i why multiply by -2i/-2i instead?
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    (Original post by timebent)
    for something like

    \dfrac{3-5i}{2i}

    why not multiply by 2i why multiply by -2i instead?
    While the conjugate of 2i is -2i, it is not necessary. As you've suggested you can multiply by \dfrac{2i}{2i}=1

    But where the complex number in the denominator has both a real an imaginary part, you multiply by the conjugate which is a-bi for a complex number a+bi) as you make use of the difference of two squares (a^2-b^2=(a-b)(a+b) \Rightarrow (a-bi)(a+bi)=a^2-(bi)^2=a^2+b^2).
    This makes the denominator real and is similar to rationalizing the denominator with surds.
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    (Original post by timebent)
    for something like

    \dfrac{3-5i}{2i}

    why not multiply by 2i/2i why multiply by -2i/-2i instead?
    Because -2i is the conjugate. Remember the conjugate of a+bi is a-bi
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    (Original post by timebent)
    for something like

    \dfrac{3-5i}{2i}

    why not multiply by 2i/2i why multiply by -2i/-2i instead?
    You can if you want to, infact, I'd skip the 2's and multiply by \frac{i}{i} myself. The importance of keeping the -bi in other examples is as KTA's answer.
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    (Original post by Kvothe the arcane)
    While the conjugate of 2i is -2i, it is not necessary. As you've suggested you can multiply by \dfrac{2i}{2i}=1

    But where the complex number in the denominator has both a real an imaginary part, you multiply by the conjugate which is a-bi for a complex number a+bi) as you make use of the difference of two squares (a^2-b^2=(a-b)(a+b) \Rightarrow (a-bi)(a+bi)=a^2-(bi)^2=a^2+b^2).
    This makes the denominator real and is similar to rationalizing the denominator with surds.
    ok thanks
    (Original post by RDKGames)
    Because -2i is the conjugate. Remember the conjugate of a+bi is a-bi
    so just because it's a conjugate that's why i'd use it?
    (Original post by Zacken)
    You can if you want to, infact, I'd skip the 2's and multiply by \frac{i}{i} myself. The importance of keeping the -bi in other examples is as KTA's answer.
    so if i multiply by just i
    then
    i get eventually -\dfrac{5}{2} - \dfrac{3}{2} i
    so am i trying to get rid of the i on the bottom or is it something else i'm trying to get rid of or change
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    (Original post by timebent)
    so am i trying to get rid of the i on the bottom or is it something else i'm trying to get rid of or change
    You just want to get rid of the i's at the bottom so your final complex number is of the form x + iy.
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    (Original post by Zacken)
    You just want to get rid of the i's at the bottom so your final complex number is of the form x + iy.
    ok thanks, you're so smart
    Just like RDK, B_ with loads of number i can't remember and Kvothe
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    (Original post by timebent;[url="tel:66513416")
    66513416[/url]]ok thanks

    so just because it's a conjugate that's why i'd use it?

    Yes because conjugates get rid off the imaginary parts. But as Zacken said, you can just multiply by i/i because the denominator has no real part to it. If the denominator is in the form a+bi where a and b =\= 0 then you need to multiply the numerator and denominator by the conjugate and multiplying them only by i won't get rid off the imaginary part in the denominator. Just treat these in the same way as rationalising the denominator with surds
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    (Original post by RDKGames)
    Yes because conjugates get rid off the imaginary parts. But as Zacken said, you can just multiply by i/i because the denominator has no real part to it. If the denominator is in the form a+bi where a and b =\= 0 then you need to multiply the numerator and denominator by the conjugate and multiplying them only by i won't get rid off the imaginary part in the denominator. Just treat these in the same way as rationalising the denominator with surfs
    ok thanks
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    Are you learning FP1 and are you in year 11, going into year 12?


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    (Original post by Chittesh14)
    Are you learning FP1 and are you in year 11, going into year 12?


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    eyy old threads
    i'm in year year 12 going to 13
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    (Original post by timebent)
    eyy old threads
    i'm in year year 12 going to 13
    Old threads ?

    Oh right, so are you doing Further Maths AS?
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    (Original post by Chittesh14)
    Old threads ?

    Oh right, so are you doing Further Maths AS?
    1 day old

    No i should've done it but i didn't take it
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    (Original post by timebent)
    1 day old

    No i should've done it but i didn't take it
    Oh right xD. Anyway, enjoy! I'm at the same stage, so let's collaborate on questions instead of asking directly on threads (because I practically do the same thing)?


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    (Original post by Chittesh14)
    Oh right xD. Anyway, enjoy! I'm at the same stage, so let's collaborate on questions instead of asking directly on threads (because I practically do the same thing)?


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    right i'm learning from new, so currently i've gone through(supposedly) the first chapter of complex numbers, i made another thread on it xD
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    (Original post by timebent)
    right i'm learning from new, so currently i've gone through(supposedly) the first chapter of complex numbers, i made another thread on it xD
    Don't worry, I've finished chapter 3. We are both learning from new. I've forgot a bit of chapter 1 so I will go back. I've forgot chapter 2 I think lol and chapter 3 was just ****ing hard (a bit)


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    (Original post by Chittesh14)
    Don't worry, I've finished chapter 3. We are both learning from new. I've forgot a bit of chapter 1 so I will go back. I've forgot chapter 2 I think lol and chapter 3 was just ****ing hard (a bit)


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    Ah check this thread, i need help on...
    http://www.thestudentroom.co.uk/show....php?t=4222072
 
 
 
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