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Math Help - complex num to cartesian form

  1. #1
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    Question complex num to cartesian form

    Hi

    How do i express a complex number in exact cartesian form?
    for example:

    (sqrt(3)+i)/((1-i)(sqrt(3)-i))

    Whats the process?

    thanks
    phil
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  2. #2
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    Quote Originally Posted by taurus View Post
    Hi

    How do i express a complex number in exact cartesian form?
    for example:

    (sqrt(3)+i)/((1-i)(sqrt(3)-i))

    Whats the process?

    thanks
    phil
    Expand the denominator so you get a number of the form a + ib. Then multiply numerator and denominator by the complex conjugate a - ib. Expand numertaor and denominator. Simplify.
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  3. #3
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    how do u expand the denominator because i get the following which is wrong:

    sqrt(3) - i + sqrt(3)i - 1
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    Quote Originally Posted by taurus View Post
    how do u expand the denominator because i get the following which is wrong:

    sqrt(3) - i - sqrt(3)i - 1
    = (\sqrt{3} - 1) - i (\sqrt{3} + 1).

    Note the correction (in red) of a careless error.
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  5. #5
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    Question

    and so the comples conjugate is:
    = (\sqrt{3} - 1) + i (\sqrt{3} + 1)

    as i just have to change the sign correct?

    so i would get:

    (\sqrt{3} + i) (\sqrt{3} - 1) + i (\sqrt{3} + 1)
    ----------------------------------
    (\sqrt{3} - 1) - i (\sqrt{3} + 1)

    How do i simplify that now?
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  6. #6
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    Hello,

    Quote Originally Posted by taurus View Post
    and so the comples conjugate is:
    = (\sqrt{3} - 1) + i (\sqrt{3} + 1)

    as i just have to change the sign correct?

    so i would get:

    (\sqrt{3} + i) (\sqrt{3} - 1) + i (\sqrt{3} + 1)
    ----------------------------------
    (\sqrt{3} - 1) - i (\sqrt{3} + 1)

    How do i simplify that now?
    No, you have to multiply the denominator and the numerator by the conjugate !

    So in the denominator, you will have (\sqrt{3}-1)^2+(\sqrt{3}+1)^2=\dots=4


    \frac{\sqrt{3}+i}{(\sqrt{3} - 1) - i (\sqrt{3} + 1)}=\frac{(\sqrt{3}+i)((\sqrt{3} - 1)+i (\sqrt{3} + 1))}{4}
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  7. #7
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    Question

    heres another example i tried but looking at answer its not quite right:

    ((1-2i)/(3+4i)) - ((2+i)/5i)
    so i first did subtraction of the top and bottom and got:
    (-1-3i)/(3-1i)
    I then performed division and got this:
    ((-3+3)/(9+1)) + ((-9+1)/(9+1))i
    which simplifies to this:
    0 + (-8/10)i
    => -4/5i

    But that isint correct as the answer is -2/5

    what am i doing wrong?
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  8. #8
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    The question you gave was \frac{1-2i}{3+4i} - \frac{2+i}{5i}

    This is two fractions with different denominators. If you recall your basic addition and subtraction of fractions, you don't subtract top and bottom. You need to make a common denominator first, and then subtract the top

    \frac{(1-2i)(5i)}{(3+4i)(5i)} - \frac{(2+i)(3+4i)}{(5i)(3+4i)} \Rightarrow \frac{(1-2i)(5i)-(2+i)(3+4i)}{(3+4i)(5i)}

    Expand that and factorise, and you should get the answer.
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