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Math Help - Imaginary numbers/complex numbers

  1. #1
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    Imaginary numbers/complex numbers

    Hi there,

    Could somebody help me with these questions, explaining the solutions and maybe providing some good references for learning?

    1)i) Simplify

    i^7, i^-3, i^9

    ii) Find
    (3 + 5i) + (7 - i)

    iii) Find
    2 / 1 - i

    2)i) Evaluate
    (2 + 3i) + (2 - 3i)

    ii)
    7 - 3i/2 + 4i

    iii)
    i^5(1 + i^5)


    I'd be very appreciative with some help in understanding/solving these types of question.

    Regards,
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  2. #2
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    Quote Originally Posted by rel85 View Post

    1)i) Simplify

    i^7, i^{-3}, i^9

    i = \sqrt{-1}

    Therefore

    i \times i = i^2= \sqrt{-1} \times \sqrt{-1} = -1

    i \times i \times i= i^3= \sqrt{-1} \times \sqrt{-1} \times \sqrt{-1}= -\sqrt{-1} = -i

    i \times i \times i \times i = i^4= (i^2)^2 = (-1)^{2} = 1

    and so on

    Quote Originally Posted by rel85 View Post


    ii) Find
    (3 + 5i) + (7 - i)
    Group like terms

    3+7 +5i-i = 10+4i

    The rest of your problems use similar ideas.
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  3. #3
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    Memorize this table.
    \begin{gathered}  i^{ - 1}  =  - i \hfill \\  i^0  = 1 \hfill \\<br />
  i^1  = i \hfill \\  i^2  =  - 1 \hfill \\  i^3  =  - i \hfill \\ \end{gathered}
    Once you have done that completely, then powers are very easy.

    If n\ge 0 then consider i^n.
    Divide n by 4 and take the remainder j then i^n = i^j. You have memorized that value.
    Example: i^{35}=i^3=-i because the remainder of 35 divided by 4 is 3.

    Now what if n<0 such as i^{-19}.
    Change that as follows: i^{-19}=\left(i^{-1}\right)^{19}=\left(-i\right)^{19}=-(-i)=i .

    Another example: i^{-10}=(-i)^{10}=i^2=-1

    BTW: If you have more than one character in an exponent, set off the whole exponent in braces.
    [tex]i^{-3}[/tex] gives i^{-3} instead of  i^-3 .
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  4. #4
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    i)

    What you should remember about powers of i is that i to the powr of a multiple of 4 is 1. so
    for example

    i^{73}=(i^{72})(i^1)=(i^{4.18})(i)=1.i=i
    so
    i^9=(i^8)(i)=(i^{4.2})(i)=1.i=i
    and
    i^{-9}=\frac{1}{i^9}=\frac{1}{i}=(\frac{1}{i})(\frac{i  }{i})=\frac{i}{i^2}=-i
    since i^2=-1
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  5. #5
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    iii)
    multiply the top and bottom of the fraction by 1+i or -1-i, then simplify

    and do the same with 2 ii) and see what you get
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  6. #6
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    Quote Originally Posted by Krahl View Post
    iii)
    multiply the top and bottom of the fraction by 1+i or -1-i, then simplify
    and do the same with 2 ii) and see what you get
    It is a lot easier than that.
    Just learn on simple fact: \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}.

    Example:  \frac{1}{{3 - 4i}} = \frac{{3 + 4i}}{25}

    So \frac{{1 + i}}<br />
{{3 - 4i}} = \frac{{\left( {1 + i} \right)\left( {3 + 4i} \right)}}<br />
{25}
    Last edited by Plato; August 25th 2009 at 11:38 AM.
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  7. #7
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    Quote Originally Posted by Plato View Post
    It is a lot easier than that.
    Just learn on simple fact: \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}.

    Example:  \frac{1}{{3 - 4i}} = \frac{{3 + 4i}}{5}

    So \frac{{1 + i}}<br />
{{3 - 4i}} = \frac{{\left( {1 + i} \right)\left( {3 + 4i} \right)}}<br />
{5}
    Thanks guys for the input so far.

    So I have:

    1. i^7 = 7/4 = 1.75 - 1 x 4 = -i

    i^{-3} = (i^{-1})^3 = (-i)^3 = -(-i) = i

    i^9 = 9/4 = 2.25 - 2 = 0.25 x 4 = i^1 = i

    I understand ii), but I the fraction puts me off, where does the 5 come from Plato? I feel I need further instruction on this and also does Q2 require the same methods?

    Regards,
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  8. #8
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    Quote Originally Posted by rel85 View Post
    I understand ii), but I the fraction puts me off, where does the 5 come from Plato?
    Do you understand the absolute value of a complex number?
    \left| {a + bi} \right| = \sqrt {a^2  + b^2 }
    So \left| {3 -4i} \right|= \sqrt {9  + 16 }=5.

    Now take the question \frac{2}{1-i}
    Because \left| {1-i} \right|=\sqrt{2} then \frac{2}{1-i}=\frac{2(1+i)}{2}=1+i.
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