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Math Help - Need Help Simplifying a Few Problems! <Imaginary Numbers>

  1. #1
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    Need Help Simplifying a Few Problems! <Imaginary Numbers>

    I tried I got the first one I think but i is equal to the square root of -1 so I'm honestly not too sure if it's fully simplified. As for the others I, don't have much of a clue at all.

    7 . 9i^26 - 14i^27 / i^7 = 9i^19 - 14i^20


    8. 3 - i / 3 + 2i


    9. -i (-3i^3 + 4i^2 - 2i + 5) = 3i^4 - 4i^3 + 2i^2 - 5i


    10. 2√(-27) - 4√(-75)
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    Re: Need Help Simplifying a Few Problems! <Imaginary Numbers>

    Hey SamBarkley2000.

    If i^2 = -1 you can simplify a great deal.

    Hint: For Q8 try multiplying by complex conjugate of denominator in the form of (3+2i)/(3+2i) [since it equals 1 it won't change the answer]
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    Re: Need Help Simplifying a Few Problems! <Imaginary Numbers>

    Quote Originally Posted by SamBarkley2000 View Post
    I tried I got the first one I think but i is equal to the square root of -1 so I'm honestly not too sure if it's fully simplified. As for the others I, don't have much of a clue at all.
    7 . 9i^26 - 14i^27 / i^7 = 9i^19 - 14i^20
    8. 3 - i / 3 + 2i
    9. -i (-3i^3 + 4i^2 - 2i + 5) = 3i^4 - 4i^3 + 2i^2 - 5i
    10. 2√(-27) - 4√(-75)
    Learn these so you can use them in your sleep.
    \begin{align*}  i^0 &= 1 \\  i^1 &=i  \\i^2 &= -1\\i^3&=-i  \end{align*}

    i^{19}=(i^4)^4(i^3)=(1)(i^3)=-i.

    If you divide 20 by 4 the remainder is 0. Thus i^{20}=i^0=1

    i^{74}=-1 because 74 divided by 4 has a remainder of 2.
    and i^2=-1.

    Also \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}
    Using the above:
    \frac{3-i}{3+2i}=\frac{(3-i)(3-2i)}{9+4}.
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    Re: Need Help Simplifying a Few Problems! <Imaginary Numbers>

    7. -23

    8. 7/13 - 9i/13

    9. 1-i

    10. -14isqrt(3)

    would those be correct chiro, thanks for the hint by the way it helped me out
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