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Math Help - Complex number question.

  1. #1
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    Complex number question.

    Find the real and imaginary parts of (-2+2i)^20(1+isquareRt(3))

    I don't really know how to solve this question, but I've tried and I haven't actually got a complex number. Could anybody tell me where I've gone wrong if it I've just done it the wrong way altogether. Sorry it looks quite messy.

    (-2+2i)^20

    sqroot(8) (cos(pi/4) + isin(pi/4)) -- found modulus and principle argument
    sqroot(8)^20 (cos(20pi/4) + isin(20pi/4)) -- multiply modulus and principle argument to power of 20
    sqroot(8)^20 (cos(pi) + isin(pi)) -- subtract principle argument by 360 until 0 < pi < 360

    sqr(8)^20 (-1 + 0) = -(sqr(8)^20) -- find cos + sin values)


    (1+iSqr(3))^12

    2(cos(pi/3) + isin(pi/3))
    4096(cos(12pi/3) + isin(12pi/3))
    4096(cos(2pi) + isin(2pi))
    4096(1 + 0)

    4096 . -(sqr(8)^20) = -4.39 x 10^12?
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  2. #2
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    Quote Originally Posted by scofield131 View Post
    Find the real and imaginary parts of (-2+2i)^20(1+isquareRt(3))

    I don't really know how to solve this question, but I've tried and I haven't actually got a complex number. Could anybody tell me where I've gone wrong if it I've just done it the wrong way altogether. Sorry it looks quite messy.

    (-2+2i)^20

    sqroot(8) (cos(pi/4) + isin(pi/4)) -- found modulus and principle argument
    sqroot(8)^20 (cos(20pi/4) + isin(20pi/4)) -- multiply modulus and principle argument to power of 20
    sqroot(8)^20 (cos(pi) + isin(pi)) -- subtract principle argument by 360 until 0 < pi < 360

    sqr(8)^20 (-1 + 0) = -(sqr(8)^20) -- find cos + sin values)


    (1+iSqr(3))^12

    2(cos(pi/3) + isin(pi/3))
    4096(cos(12pi/3) + isin(12pi/3))
    4096(cos(2pi) + isin(2pi))
    4096(1 + 0)

    4096 . -(sqr(8)^20) = -4.39 x 10^12?
    Convert them to polars:

    |-2 + 2i| = \sqrt{(-2)^2 + 2^2} = 2\sqrt{2}

    \arg{(-2 + 2i)} = \pi - \arctan{\frac{2}{2}} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.


    |1 + \sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2.

    \arg{(1 + \sqrt{3}i)} = \arctan{\frac{\sqrt{3}}{1}} = \frac{\pi}{3}.


    Therefore:

    (-2 + 2i)^{20}(1 + \sqrt{3}i) = \left(2\sqrt{2}e^{\frac{3\pi}{4}i}\right)^{20}\lef  t(2e^{\frac{\pi}{3}i}\right)

     = \left[\left(2\sqrt{2}\right)^{20}e^{15i}\right]\left(2e^{\frac{\pi}{3}i}\right)

     = 2^{31}e^{15\pi i + \frac{\pi}{3}i}

     = 2^{31}e^{\frac{46 \pi}{3}i}

     = 2^{31}\left[\cos{\left(\frac{46\pi}{3}\right)} + i\sin{\left(\frac{46\pi}{3}\right)}\right]

     = 2^{31}\left[\cos{\left(\frac{4\pi}{3}\right)} + i\sin{\left(\frac{4\pi}{3}\right)}\right]

     = 2^{31}\left[-\cos{\left(\frac{\pi}{3}\right)} - i\sin{\left(\frac{\pi}{3}\right)}\right]

     = 2^{31}\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)

     = -2^{30} - 2^{30}\sqrt{3}i.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    Convert them to polars:

    |-2 + 2i| = \sqrt{(-2)^2 + 2^2} = 2\sqrt{2}

    \arg{(-2 + 2i)} = \pi - \arctan{\frac{2}{2}} = \pi - \frac{\pi}{4} = \frac{3\pi}{4}.


    |1 + \sqrt{3}i| = \sqrt{1^2 + (\sqrt{3})^2} = \sqrt{4} = 2.

    \arg{(1 + \sqrt{3}i)} = \arctan{\frac{\sqrt{3}}{1}} = \frac{\pi}{3}.


    Therefore:

    (-2 + 2i)^{20}(1 + \sqrt{3}i) = \left(2\sqrt{2}e^{\frac{3\pi}{4}i}\right)^{20}\lef  t(2e^{\frac{\pi}{3}i}\right)

    *  = \left[\left(2\sqrt{2}\right)^{20}<b>e^{15i}</b>\right]\left(2e^{\frac{\pi}{3}i}\right)

     = 2^{31}e^{15\pi i + \frac{\pi}{3}i}

     = 2^{31}e^{\frac{46 \pi}{3}i}

     = 2^{31}\left[\cos{\left(\frac{46\pi}{3}\right)} + i\sin{\left(\frac{46\pi}{3}\right)}\right]

     = 2^{31}\left[\cos{\left(\frac{4\pi}{3}\right)} + i\sin{\left(\frac{4\pi}{3}\right)}\right]

     = 2^{31}\left[-\cos{\left(\frac{\pi}{3}\right)} - i\sin{\left(\frac{\pi}{3}\right)}\right]

     = 2^{31}\left(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\right)

     = -2^{30} - 2^{30}\sqrt{3}i.
    Hey, thanks for helping. However, could you explain it further from the star. I'm, in particular, unsure where the 15 came from?

    Thanks.
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  4. #4
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    This is, I think, a simpler approach to this problem.
    Because \left( { - 2 + 2i} \right) = \sqrt 8 \exp \left( {\frac{{3\pi i}}{4}} \right) then  \left( { - 2 + 2i} \right)^{20}  = \left( {\sqrt 8 } \right)^{20} \exp \left( {\frac{{3\pi i}}{4}} \right)^{20}  = 8^{10} \exp \left( { 15\pi i} \right) =  - 2^{30} .
    Now just multiply -2^{30} by (1+i\sqrt{3}).
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  5. #5
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    Quote Originally Posted by Plato View Post
    This is, I think, a simpler approach to this problem.
    Because \left( { - 2 + 2i} \right) = \sqrt 8 \exp \left( {\frac{{3\pi i}}{4}} \right) then  \left( { - 2 + 2i} \right)^{20}  = \left( {\sqrt 8 } \right)^{20} \exp \left( {\frac{{3\pi i}}{4}} \right)^{20}  = 8^{10} \exp \left( { 15\pi i} \right) =  - 2^{30} .
    Now just multiply -2^{30} by (1+i\sqrt{3}).
    Thank you, that is simpler. However, call me stupid, I still don't understand why exp(3pi/4)^20 turned into 15pi? As when I do 135^20 I get 4.04 x 10^42, not 15pi?
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  6. #6
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    Quote Originally Posted by scofield131 View Post
    Thank you, that is simpler. However, call me stupid, I still don't understand why exp(3pi/4)^20 turned into 15pi? As when I do 135^20 I get 4.04 x 10^42, not 15pi?
    Recall that \left[ {r\exp (\phi i)} \right]^n  = r^n \exp (n\phi i)<br />

    \left( {\frac{{3\pi i}}<br />
{4}} \right)\left( {20} \right) = 15\pi i,\quad \left(\frac{3}<br />
{4}\right)(20) = 15
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  7. #7
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    Quote Originally Posted by Plato View Post
    Recall that \left[ {r\exp (\phi i)} \right]^n  = r^n \exp (n\phi i)<br />

    \left( {\frac{{3\pi i}}<br />
{4}} \right)\left( {20} \right) = 15\pi i,\quad \left(\frac{3}<br />
{4}\right)(20) = 15
    Thank you, I get it now. Sorry to keep bothering you, but could you perhaps explain the rest of the answer too? In particular, how you got to -2^30? And then if I then go back to the start and do the same for \left( { 1 + i } \right)^{12} " alt="sqrt{3}} \right)^{12} " /> then multiply them together?
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  8. #8
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    Quote Originally Posted by Plato View Post
    Recall that \left[ {r\exp (\phi i)} \right]^n  = r^n \exp (n\phi i)<br />

    \left( {\frac{{3\pi i}}<br />
{4}} \right)\left( {20} \right) = 15\pi i,\quad \left(\frac{3}<br />
{4}\right)(20) = 15
    Thank you, I get it now. Sorry to keep bothering you, but could you perhaps explain the rest of the answer too? In particular, how you got to -2^30? And then if I then go back to the start and do the same for \left( { 1 + i sqrt{3}} \right)^{12}

    Then multply them together?
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  9. #9
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    \left( 8 \right)^{10}-\left( 2^3 \right)^{10}=2^{30}
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