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Math Help - complex numbers

  1. #1
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    complex numbers

    Hi all,

    I am having trouble getting the answer to:
    [(1-sqrt(3)i)/ (sqrt(3) + i)]^7

    where i = complex number

    Thanks in advance,
    ArTiCk
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  2. #2
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    Quote Originally Posted by ArTiCK View Post
    Hi all,

    I am having trouble getting the answer to:
    [(1-sqrt(3)i)/ (sqrt(3) + i)]^7

    where i = complex number

    Thanks in advance,
    ArTiCk
    My advice is to first convert the numbers into polar form.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    My advice is to first convert the numbers into polar form.
    I have already converted the numbers into polar form...
    i have it = e^(-7i*pi/3)/ [e^(7i*pi/6)]

    The trouble i have is that i tried converting e^(-7i*pi/3) into Cartesian form as well as the denominator... but i am having trouble with the nominator. What i did was:

    e^(-7i*pi/3) = cos(-7*pi/3) + i*sin(-7*pi/3) = cos (-2*pi - pi/3) + i*sin(-2*pi -pi/3) = cos(-pi/3) + i*sin(-pi/3)

    I think it is there where i have gone wrong, help would be appreciated

    Thanks, ArTiCk
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  4. #4
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    Hello, ArTiCK!

    Simplify: .  \left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7
    Ratinalize the "inside" first . . .

    . . \frac{1-i\sqrt{3}}{\sqrt{3} + i}\cdot\frac{\sqrt{3}-i}{\sqrt{3}-i} \;=\;\frac{\sqrt{3} - i - 3i +i^2\sqrt{3}}{3-i^2} \;=\;\frac{\sqrt{3} -4i-\sqrt{3}}{3+1} \;=\;\frac{-4i}{4} \;=\;-i


    So we have: . (-i)^7 \;=\; i

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  5. #5
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    Quote Originally Posted by ArTiCK View Post
    I have already converted the numbers into polar form...
    i have it = e^(-7i*pi/3)/ [e^(7i*pi/6)]

    The trouble i have is that i tried converting e^(-7i*pi/3) into Cartesian form as well as the denominator... but i am having trouble with the nominator. What i did was:

    e^(-7i*pi/3) = cos(-7*pi/3) + i*sin(-7*pi/3) = cos (-2*pi - pi/3) + i*sin(-2*pi -pi/3) = cos(-pi/3) + i*sin(-pi/3)

    I think it is there where i have gone wrong, help would be appreciated

    Thanks, ArTiCk
    \left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7 = <br />
\left[ \frac{2 \, \text{cis}  \left( -\frac{\pi}{3}\right) }{2 \, \text{cis}  \left( \frac{\pi}{6}\right)} \right]^7 = \left[ \text{cis} \left( -\frac{\pi}{3} - \frac{\pi}{6} \right)\right]^7 = \left[ \text{cis} \left( -\frac{\pi}{2} \right)\right]^7 = (-i)^7.
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  6. #6
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    Hello, Mr. F !


    Quote Originally Posted by mr fantastic View Post

    \left(\frac{1-i\sqrt{3}}{\sqrt{3}+i}\right)^7 = <br />
\left[ \frac{2 \, \text{cis}  \left( -\frac{\pi}{3}\right) }{2 \, \text{cis}  \left( \frac{\pi}{6}\right)} \right]^7 = \left[ \text{cis} \left( -\frac{\pi}{3} - \frac{\pi}{6} \right)\right]^7 = \left[ \text{cis} \left( -\frac{\pi}{2} \right)\right]^7 = (-i)^7.

    Lovely!

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