Results 1 to 8 of 8

Thread: Imaginary numbers/complex numbers

  1. #1
    Newbie
    Joined
    Aug 2009
    From
    United Kingdom
    Posts
    17

    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

    $\displaystyle i^7, i^-3, i^9$

    ii) Find
    $\displaystyle (3 + 5i) + (7 - i)$

    iii) Find
    $\displaystyle 2 / 1 - i$

    2)i) Evaluate
    $\displaystyle (2 + 3i) + (2 - 3i)$

    ii)
    $\displaystyle 7 - 3i/2 + 4i$

    iii)
    $\displaystyle i^5(1 + i^5)$


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

    Regards,
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,237
    Thanks
    33
    Quote Originally Posted by rel85 View Post

    1)i) Simplify

    $\displaystyle i^7, i^{-3}, i^9$

    $\displaystyle i = \sqrt{-1} $

    Therefore

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

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

    $\displaystyle 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
    $\displaystyle (3 + 5i) + (7 - i)$
    Group like terms

    $\displaystyle 3+7 +5i-i = 10+4i$

    The rest of your problems use similar ideas.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,774
    Thanks
    2823
    Awards
    1
    Memorize this table.
    $\displaystyle \begin{gathered} i^{ - 1} = - i \hfill \\ i^0 = 1 \hfill \\
    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 $\displaystyle n\ge 0$ then consider $\displaystyle i^n$.
    Divide $\displaystyle n$ by 4 and take the remainder $\displaystyle j$ then $\displaystyle i^n = i^j$. You have memorized that value.
    Example: $\displaystyle i^{35}=i^3=-i$ because the remainder of 35 divided by 4 is 3.

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

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

    BTW: If you have more than one character in an exponent, set off the whole exponent in braces.
    [tex]i^{-3}[/tex] gives $\displaystyle i^{-3}$ instead of $\displaystyle i^-3 $.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2009
    Posts
    193
    Thanks
    5
    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

    $\displaystyle i^{73}=(i^{72})(i^1)=(i^{4.18})(i)=1.i=i$
    so
    $\displaystyle i^9=(i^8)(i)=(i^{4.2})(i)=1.i=i$
    and
    $\displaystyle i^{-9}=\frac{1}{i^9}=\frac{1}{i}=(\frac{1}{i})(\frac{i }{i})=\frac{i}{i^2}=-i$
    since i^2=-1
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jul 2009
    Posts
    193
    Thanks
    5
    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
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,774
    Thanks
    2823
    Awards
    1
    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: $\displaystyle \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$.

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

    So $\displaystyle \frac{{1 + i}}
    {{3 - 4i}} = \frac{{\left( {1 + i} \right)\left( {3 + 4i} \right)}}
    {25}$
    Last edited by Plato; Aug 25th 2009 at 11:38 AM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Aug 2009
    From
    United Kingdom
    Posts
    17
    Quote Originally Posted by Plato View Post
    It is a lot easier than that.
    Just learn on simple fact: $\displaystyle \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}$.

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

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

    So I have:

    1. $\displaystyle i^7 = 7/4 = 1.75 - 1 x 4 = -i$

    $\displaystyle i^{-3} = (i^{-1})^3 = (-i)^3 = -(-i) = i$

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

    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,
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,774
    Thanks
    2823
    Awards
    1
    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?
    $\displaystyle \left| {a + bi} \right| = \sqrt {a^2 + b^2 } $
    So $\displaystyle \left| {3 -4i} \right|= \sqrt {9 + 16 }=5$.

    Now take the question $\displaystyle \frac{2}{1-i}$
    Because $\displaystyle \left| {1-i} \right|=\sqrt{2}$ then $\displaystyle \frac{2}{1-i}=\frac{2(1+i)}{2}=1+i$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Sep 27th 2010, 03:14 PM
  2. Replies: 2
    Last Post: Dec 9th 2009, 03:14 PM
  3. Complex/imaginary numbers?
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: Aug 20th 2009, 04:59 PM
  4. Replies: 1
    Last Post: May 24th 2007, 03:49 AM
  5. Complex Numbers- Imaginary numbers
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jan 24th 2007, 12:34 AM

Search Tags


/mathhelpforum @mathhelpforum