Results 1 to 9 of 9

Thread: Complex number question.

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    25

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    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:

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

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


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

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


    Therefore:

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

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

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

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

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

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

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

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

    $\displaystyle = -2^{30} - 2^{30}\sqrt{3}i$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    25
    Quote Originally Posted by Prove It View Post
    Convert them to polars:

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

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


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

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


    Therefore:

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

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

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

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

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

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

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

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

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

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    This is, I think, a simpler approach to this problem.
    Because $\displaystyle \left( { - 2 + 2i} \right) = \sqrt 8 \exp \left( {\frac{{3\pi i}}{4}} \right)$ then $\displaystyle \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 $\displaystyle -2^{30}$ by $\displaystyle (1+i\sqrt{3})$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2009
    Posts
    25
    Quote Originally Posted by Plato View Post
    This is, I think, a simpler approach to this problem.
    Because $\displaystyle \left( { - 2 + 2i} \right) = \sqrt 8 \exp \left( {\frac{{3\pi i}}{4}} \right)$ then $\displaystyle \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 $\displaystyle -2^{30}$ by $\displaystyle (1+i\sqrt{3})$.
    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?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    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 $\displaystyle \left[ {r\exp (\phi i)} \right]^n = r^n \exp (n\phi i)
    $

    $\displaystyle \left( {\frac{{3\pi i}}
    {4}} \right)\left( {20} \right) = 15\pi i,\quad \left(\frac{3}
    {4}\right)(20) = 15$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2009
    Posts
    25
    Quote Originally Posted by Plato View Post
    Recall that $\displaystyle \left[ {r\exp (\phi i)} \right]^n = r^n \exp (n\phi i)
    $

    $\displaystyle \left( {\frac{{3\pi i}}
    {4}} \right)\left( {20} \right) = 15\pi i,\quad \left(\frac{3}
    {4}\right)(20) = 15$
    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 $\displaystyle \left( { 1 + i$$\displaystyle sqrt{3}} \right)^{12} $ then multiply them together?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Nov 2009
    Posts
    25
    Quote Originally Posted by Plato View Post
    Recall that $\displaystyle \left[ {r\exp (\phi i)} \right]^n = r^n \exp (n\phi i)
    $

    $\displaystyle \left( {\frac{{3\pi i}}
    {4}} \right)\left( {20} \right) = 15\pi i,\quad \left(\frac{3}
    {4}\right)(20) = 15$
    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 $\displaystyle \left( { 1 + i sqrt{3}} \right)^{12} $

    Then multply them together?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1
    $\displaystyle \left( 8 \right)^{10}-\left( 2^3 \right)^{10}=2^{30}$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. complex number question..
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Jul 2nd 2011, 12:42 PM
  2. Complex Number Question
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Apr 7th 2010, 10:14 PM
  3. Another complex number question:
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Dec 16th 2009, 02:12 AM
  4. complex number question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Jul 27th 2009, 07:02 PM

Search Tags


/mathhelpforum @mathhelpforum