Results 1 to 5 of 5

Thread: i evaluations

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

    i evaluations

    Hi,

    I'd be very appreciative if someone could explain the steps and methodology to 'evaluate' the following:

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

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

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

    Many regards,
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    May 2009
    Posts
    471
    Quote Originally Posted by rel85 View Post
    Hi,

    I'd be very appreciative if someone could explain the steps and methodology to 'evaluate' the following:

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

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

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

    Many regards,
    i) $\displaystyle (2 + 3i)(2 - 3i)=2(2)+2(-3i)+3i(2)+3i(-3i)$ we just foiled
    $\displaystyle =4-6i+6i-9i^2=4-9(-1)=4+9=13$ and since $\displaystyle i=\sqrt{-1}, i^2=-1$



    Should there be parenthesis for the fraction in ii)? otherwise just combine the i terms



    iii) $\displaystyle i^5(1 + i^5)$
    $\displaystyle i^5=i^4*i=i^2*i^2*i=(-1)(-1)i=i$ so we now have

    $\displaystyle i(1+i)=i+i^2=i-1=-1+i$

    In general, to evaluate $\displaystyle i^n$ divide n by 4, and take the remainder as the power, for instance $\displaystyle i^{27}$ $\displaystyle \frac{27}{4}=6$ remainder 3, so $\displaystyle i^{27}=i^3=-i$

    $\displaystyle i^0=1$
    $\displaystyle i^1=i$
    $\displaystyle i^2=-1$
    $\displaystyle i^3=-i$
    $\displaystyle i^4=1$ this is why we divide by 4, the cycle repeats as you see here
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    From
    United Kingdom
    Posts
    17
    Quote Originally Posted by artvandalay11 View Post
    i) $\displaystyle (2 + 3i)(2 - 3i)=2(2)+2(-3i)+3i(2)+3i(-3i)$ we just foiled
    $\displaystyle =4-6i+6i-9i^2=4-9(-1)=4+9=13$ and since $\displaystyle i=\sqrt{-1}, i^2=-1$



    Should there be parenthesis for the fraction in ii)? otherwise just combine the i terms



    iii) $\displaystyle i^5(1 + i^5)$
    $\displaystyle i^5=i^4*i=i^2*i^2*i=(-1)(-1)i=i$ so we now have

    $\displaystyle i(1+i)=i+i^2=i-1=-1+i$

    In general, to evaluate $\displaystyle i^n$ divide n by 4, and take the remainder as the power, for instance $\displaystyle i^{27}$ $\displaystyle \frac{27}{4}=6$ remainder 3, so $\displaystyle i^{27}=i^3=-i$

    $\displaystyle i^0=1$
    $\displaystyle i^1=i$
    $\displaystyle i^2=-1$
    $\displaystyle i^3=-i$
    $\displaystyle i^4=1$ this is why we divide by 4, the cycle repeats as you see here
    Am I doing something wrong for the first one?
    I get $\displaystyle 4 - 9i^2 + 6^i - 6^i$ from expanding the brackets which becomes $\displaystyle 4 - 9i^2 = 4 - 9(-1) == 13$

    Does it stop there, where is the i^2 = -1 coming from at this point?


    No parenthesises for the second, just a fraction... Group like terms how?

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2009
    Posts
    193
    Thanks
    5
    $\displaystyle

    7 - 3i / 2 + 4i
    $

    $\displaystyle 7-\frac{3}{2}i+4i$

    $\displaystyle 7+(4-\frac{3}{2})i$

    group the i terms as you would x terms
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    May 2009
    Posts
    471
    $\displaystyle (\sqrt{x})^2=x$

    $\displaystyle i=\sqrt{-1}$ so square both sides

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


    $\displaystyle 7-\frac{3i}{2}+4i=7-\frac{3i}{2}+\frac{8i}{2}$

    $\displaystyle =7+\frac{5i}{2}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. base integrations / evaluations (calc2)
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Feb 21st 2009, 09:37 AM

Search Tags


/mathhelpforum @mathhelpforum