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Math Help - More complex number questions

  1. #1
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    More complex number questions

    Can help me to solve this?

    4. If z=x+yi and z^2=a+bi,x,y,a,b \in IR, prove that
    2x^2= \sqrt {(a^2+b^2)}+a
    By solving the equation z^4+6z^2+25=0 for z^2 or otherwise, express each root of the equation in the form x+yi.

    5a. Express the complex numbers 1+i and 1-i in the form r(cos \theta + i sin \theta ), where r is the modulus and  \theta the argument of  \frac {(1+i)^5}{(1-i)^7}.

    5b. Given that z_{1}=2+i and z_{2}=-2+4i, find, in the form a+bi, the complex number z such that  \frac {1}{z}= \frac {1}{z_{1}}+ \frac {1}{z_{2}}.
    Find the modulus and argument of z, giving your answer for arg z in the range - \pi <argz \leq \pi, correct to 3 decimal places.


    Answers:

    4. \pm(1+2i),\pm(1-2i)
    5a. \sqrt{2}(cos\frac{\pi}{4}+i sin \frac{\pi}{4}), \sqrt{2}[cos(-\frac{\pi}{4})+i sin(-\frac{\pi}{4})]; \frac{1}{2},\pi
    5b. \frac{6}{5}+\frac{8}{5}i; 2, 0.927 rad
    Last edited by cloud5; June 19th 2009 at 05:26 AM. Reason: Moved questions from original thread
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  2. #2
    Senior Member I-Think's Avatar
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    No.1

    Ans for NO.1

    z=x+yi, z^2=x^2-y^2+2xyi....(1), z^2=a+bi...(2)
    Hence, a=x^2-y^2...(3)
    (from 2) \bmod{z^2}=\sqrt{a^2+b^2},
    (from 1) \bmod{z^2}=\sqrt{(x^2-y^2)^2+)2xy)^2)} =\sqrt{x^4-2x^2y^2+y^4+4x^2y^2} =\sqrt{x^4+2x^2y^2+y^4}=x^2+y^2...(4)

    a=x^2-y^2
    \sqrt{a^2+b^2}=x^2+y^2
    Add these two equations
    \sqrt{a^2+b^2}+a=2x^2


    Solve for z^2 by quadratic formula

    z^2=\frac{-6\pm\sqrt{36-100}}{2}=-3\pm4i

    Set z= c+di
    z^2=-3+4i or z^2=-3-4i

    z^2=c^2-d^2+2cdi
    Hence c^2-d^2=-3
     2cd=\pm4

    d=\pm\frac{2}{c}

    c^2-(\pm\frac{2}{c})^2=-3

    c^2-\frac{4}{c^2}=-3

    c^4+3c^2-4=0

    (c^2+4)(c^2-1)=0

    c^2=-4, c^2=1

    c=\pm2i,\pm1


    If  c=2i, If c=-2i, If c=1, If c=-1
    Then d=\pm\frac{4}{4i}, Then d=\pm\frac{4}{-4i} ,, Then d=\pm\frac{4}{2} ,, Then d=\pm\frac{4}{-2}
    d=\mp{i} ,, d=\pm{i} ,,  d=\pm2 ,, d=\mp2
    z=\pm(1+2i),\pm(1-2i)

    For number 5b, is it or z_2=2+4i?
    Last edited by I-Think; June 18th 2009 at 06:00 AM. Reason: To correct the pm signs
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  3. #3
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    Quote Originally Posted by cloud5 View Post
    5a. Express the complex numbers 1+i and 1-i in the form r(cos \theta + i sin \theta ), where r is the modulus and  \theta the argument of  \frac {(1+i)^5}{(1-i)^7}.
    Not sure what you're asking. I think it's easy to see that
    1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right),

    and that
    1 - i = \sqrt{2}\left(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4}\right).

    DeMoivre's Theorem states that if
    z = r(\cos \theta + i \sin \theta )

    then
    z^n = r^n(\cos n\theta + i \sin n\theta ).

    So
    \begin{aligned}<br />
(1 + i)^5 &= (\sqrt{2})^5 \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right) \\<br />
&= 4\sqrt{2} \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right)<br />
\end{aligned}

    and
    \begin{aligned}<br />
(1 - i)^7 &= (\sqrt{2})^7 \left(\cos \frac{49\pi}{4} + i\sin \frac{49\pi}{4}\right) \\<br />
&= 8\sqrt{2} \left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)<br />
\end{aligned}
    (49pi/4 and pi/4 are coterminal angles.)

    When you divide two complex numbers in polar form, you divide the two moduli and subtract the arguments:
    \begin{aligned}<br />
\frac {(1+i)^5}{(1-i)^7} &= \frac{4\sqrt{2} \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right)}{8\sqrt{2} \left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)} \\<br />
&= \frac{4\sqrt{2}}{8\sqrt{2}}\left[\cos \left(\frac{5\pi}{4} - \frac{\pi}{4}\right) + i\sin \left(\frac{5\pi}{4} - \frac{\pi}{4}\right)\right] \\<br />
&= \frac{1}{2}(\cos \pi + i \sin \pi) \\<br />
&= -\frac{1}{2}<br />
\end{aligned}


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  4. #4
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    Quote Originally Posted by I-Think View Post
    For number 5b, is it or z_2=2+4i?
    I can get the fraction but it contain the negative sign... Correction... z_{2}=-2+4i
    Last edited by cloud5; June 19th 2009 at 05:56 AM.
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  5. #5
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    Quote Originally Posted by yeongil View Post
    Not sure what you're asking. I think it's easy to see that
    1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right),

    and that
    1 - i = \sqrt{2}\left(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4}\right).

    DeMoivre's Theorem states that if
    z = r(\cos \theta + i \sin \theta )

    then
    z^n = r^n(\cos n\theta + i \sin n\theta ).

    So
    \begin{aligned}<br />
(1 + i)^5 &= (\sqrt{2})^5 \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right) \\<br />
&= 4\sqrt{2} \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right)<br />
\end{aligned}

    and
    \begin{aligned}<br />
(1 - i)^7 &= (\sqrt{2})^7 \left(\cos \frac{49\pi}{4} + i\sin \frac{49\pi}{4}\right) \\<br />
&= 8\sqrt{2} \left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)<br />
\end{aligned}
    (49pi/4 and pi/4 are coterminal angles.)

    When you divide two complex numbers in polar form, you divide the two moduli and subtract the arguments:
    \begin{aligned}<br />
\frac {(1+i)^5}{(1-i)^7} &= \frac{4\sqrt{2} \left(\cos \frac{5\pi}{4} + i\sin \frac{5\pi}{4}\right)}{8\sqrt{2} \left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right)} \\<br />
&= \frac{4\sqrt{2}}{8\sqrt{2}}\left[\cos \left(\frac{5\pi}{4} - \frac{\pi}{4}\right) + i\sin \left(\frac{5\pi}{4} - \frac{\pi}{4}\right)\right] \\<br />
&= \frac{1}{2}(\cos \pi + i \sin \pi) \\<br />
&= -\frac{1}{2}<br />
\end{aligned}


    01
    Can you explain all of this? and I think your answer is a bit different from the answer above or it might be my book's fault? but I need this 1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right),

    and that
    1 - i = \sqrt{2}\left(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4}\right) explaination.
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  6. #6
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    Quote Originally Posted by cloud5 View Post
    Can you explain all of this? and I think your answer is a bit different from the answer above or it might be my book's fault? but I need this 1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right),

    and that
    1 - i = \sqrt{2}\left(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4}\right) explaination.
    \frac{7\pi}{4}= \frac{2\pi}- \frac{-\pi}{4} so that cos(\frac{7\pi}{4}= cos(\frac{-\pi}{4})= cos(\frac{\pi}{4}) and sin(\frac{7\pi}{4})= sin(-\frac{\pi}{4})= -sin(\frac{\pi}{4}).
    Was that what was bothering you?
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  7. #7
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    Quote Originally Posted by cloud5 View Post
    I can get the fraction but it contain the negative sign... Correction... z_{2}=-2+4i
    I think the answer given is not correct .

    \frac{1}{z} =\frac{1}{z_1}+\frac{1}{z_2}

    =\frac{1}{2+i}+\frac{1}{-2+4i}

    =\frac{5i}{-8+6i}

    z=\frac{6}{5}-\frac{8}{5}i
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  8. #8
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    Quote Originally Posted by cloud5 View Post
    Can you explain all of this? and I think your answer is a bit different from the answer above or it might be my book's fault? but I need this 1 + i = \sqrt{2}\left(\cos \frac{\pi}{4} + i\sin \frac{\pi}{4}\right),

    and that
    1 - i = \sqrt{2}\left(\cos \frac{7\pi}{4} + i\sin \frac{7\pi}{4}\right) explaination.
    For a complex number a + bi, to convert to polar form r(\cos \theta + \sin \theta), you need to know the following:
    r = \sqrt{a^2 + b^2}
    \theta = \tan^{-1}\left(\frac{b}{a}\right) (subject to correction, since inverse tangent only gives angles in -\pi/2 < \theta < \pi/2.

    For 1 + i (or 1 + 1i),
    r = \sqrt{1^2 + 1^2} = \sqrt{2}
    \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}

    For 1 - i (or 1 - 1i),
    r = \sqrt{1^2 + (-1)^2} = \sqrt{2}
    \theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} + 2\pi = \frac{7\pi}{4}
    As HallsofIvy has shown, -\frac{\pi}{4} & \frac{7\pi}{4} are coterminal angles. I added 2\pi above because I wanted an angle that was in the interval [0, 2\pi).

    As far as I'm concerned my answers match what you originally posted.
    5a. \sqrt{2}(cos\frac{\pi}{4}+i sin \frac{\pi}{4}), \sqrt{2}[cos(-\frac{\pi}{4})+i sin(-\frac{\pi}{4})]; \frac{1}{2},\pi
    The last two numbers, \frac{1}{2} \text{and} \pi, are the modulus and argument, respectively, of my answer
    \frac{1}{2}(\cos \pi + i \sin \pi).


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  9. #9
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    There is a very simple way to work 5a.
    This is a grossly underused identity. \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}

    Notice that \left( {1 + i} \right)^5 \left( {1 + i} \right)^7  = \left( {1 + i} \right)^{12}  = \left( {{\sqrt 2 }\exp (\pi i/4)} \right)^{12}  =  - 2^6

    So \frac{{\left( {1 + i} \right)^5 }}{{\left( {1 - i} \right)^7 }} = \frac{{\left( {1 + i} \right)^{12} }}{{\left( {\sqrt 2 } \right)^{14} }} = \frac{{ - 2^6 }}{{2^7 }} = \frac{{ - 1}}{2}.
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  10. #10
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    Quote Originally Posted by HallsofIvy View Post
    \frac{7\pi}{4}= \frac{2\pi}- \frac{-\pi}{4} so that cos(\frac{7\pi}{4}= cos(\frac{-\pi}{4})= cos(\frac{\pi}{4}) and sin(\frac{7\pi}{4})= sin(-\frac{\pi}{4})= -sin(\frac{\pi}{4}).
    Was that what was bothering you?
    After looking at yeongil's explanation, I finally got the point. Thanks for the help~
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  11. #11
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    Quote Originally Posted by mathaddict View Post
    z=\frac{6}{5}-\frac{8}{5}i
    I got that too, but how to change \frac{1}{z}to z?
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  12. #12
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    Quote Originally Posted by cloud5 View Post
    I got that too, but how to change \frac{1}{z}to z?
    You mean from this:
    \frac{1}{z}=\frac{5i}{-8+6i}
    to this?
    z=\frac{6}{5}-\frac{8}{5}i

    Just take the reciprocal of both sides and multiply top and bottom by -5i:
    \begin{aligned}<br />
\frac{1}{z} &= \frac{5i}{-8+6i} \\<br />
\frac{z}{1} &= \frac{-8+6i}{5i} \\<br />
\end{aligned}
    \begin{aligned}<br />
z &= \frac{-8+6i}{5i} \times \frac{-5i}{-5i} \\<br />
&= \frac{40i + 30}{25} \\<br />
&= \frac{30}{25} + \frac{40}{25}i \\<br />
&= \frac{6}{5} + \frac{8}{5}i<br />
\end{aligned}

    Umm, mathaddict, I'm not getting what you got. In fact, this matches cloud5's answer in the original post.


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  13. #13
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    Quote Originally Posted by yeongil View Post
    You mean from this:
    \frac{1}{z}=\frac{5i}{-8+6i}
    to this?
    z=\frac{6}{5}-\frac{8}{5}i

    Just take the reciprocal of both sides and multiply top and bottom by -5i:
    \begin{aligned}<br />
\frac{1}{z} &= \frac{5i}{-8+6i} \\<br />
\frac{z}{1} &= \frac{-8+6i}{5i} \\<br />
\end{aligned}
    \begin{aligned}<br />
z &= \frac{-8+6i}{5i} \times \frac{-5i}{-5i} \\<br />
&= \frac{40i + 30}{25} \\<br />
&= \frac{30}{25} + \frac{40}{25}i \\<br />
&= \frac{6}{5} + \frac{8}{5}i<br />
\end{aligned}

    01
    Yeah! Thank you!
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