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Math Help - i evaluations

  1. #1
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    i evaluations

    Hi,

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

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

    ii) 7 - 3i / 2 + 4i

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

    Many regards,
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  2. #2
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    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) (2 + 3i)(2 - 3i)

    ii) 7 - 3i / 2 + 4i

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

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



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



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

    i(1+i)=i+i^2=i-1=-1+i

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

    i^0=1
    i^1=i
    i^2=-1
    i^3=-i
    i^4=1 this is why we divide by 4, the cycle repeats as you see here
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  3. #3
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    Quote Originally Posted by artvandalay11 View Post
    i) (2 + 3i)(2 - 3i)=2(2)+2(-3i)+3i(2)+3i(-3i) we just foiled
    =4-6i+6i-9i^2=4-9(-1)=4+9=13 and since i=\sqrt{-1}, i^2=-1



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



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

    i(1+i)=i+i^2=i-1=-1+i

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

    i^0=1
    i^1=i
    i^2=-1
    i^3=-i
    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 4 - 9i^2 +  6^i - 6^i from expanding the brackets which becomes 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!
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  4. #4
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    Thanks
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    <br /> <br />
7 - 3i / 2 + 4i<br />

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

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

    group the i terms as you would x terms
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  5. #5
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    (\sqrt{x})^2=x

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

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


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

    =7+\frac{5i}{2}
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