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Math Help - Finding roots in a + ib form

  1. #1
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    Finding roots in a + ib form

    Express the following in the form
    a + ib. Give all values and make a polar plot of
    the points or the vector vectors that represent your result.
    (a)
    (9i)^(1/2)

    is this the answer? =3*(cos pi/2 + i sin pi/2)
    =3i

    so how do we represent this on polar plot?

    (b)(-64i)^(1/4)=(64^0.25)*(cos theta/4 + i sin theta/4)

    here im getting theta= -pi and +pi so which one do we take?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by sandy View Post
    Express the following in the form
    a + ib. Give all values and make a polar plot of
    the points or the vector vectors that represent your result.
    (a)
    (9i)^(1/2)

    is this the answer? =3*(cos pi/2 + i sin pi/2)
    =3i
    obviously not. (9i)^{1/2} \ne 3i, since if you square 3i you get -9, not 9i

    Let (9i)^{1/2} = a + ib

    \Rightarrow 9i = (a + ib)^2 = a^2 - b^2 + 2abi

    Thus, a^2 - b^2 = 0 ........(1) and

    2ab = 9 ...............(2)

    Now solve this system simultaneously, and note that a,b \in \mathbb R
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    Quote Originally Posted by Jhevon View Post
    obviously not. (9i)^{1/2} \ne 3i, since if you square 3i you get -9, not 9i

    Let (9i)^{1/2} = a + ib

    \Rightarrow 9i = (a + ib)^2 = a^2 - b^2 + 2abi

    Thus, a^2 - b^2 = 0 ........(1) and

    2ab = 9 ...............(2)

    Now solve this system simultaneously, and note that a,b \in \mathbb R
    Isn't this a multi-branch function?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Isn't this a multi-branch function?
    i don't think they want to get that complicated here. it seems they are just asking for the 2 square roots.
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    Quote Originally Posted by sandy View Post
    Express the following in the form
    a + ib. Give all values and make a polar plot of
    the points or the vector vectors that represent your result.
    (a)
    (9i)^(1/2)

    is this the answer? =3*(cos pi/2 + i sin pi/2)
    =3i

    so how do we represent this on polar plot?

    (b)(-64i)^(1/4)=(64^0.25)*(cos theta/4 + i sin theta/4)

    here im getting theta= -pi and +pi so which one do we take?
    You need to remember that there are exactly two square roots, and they are evenly spaced around the unit circle. In other words, they differ by an angle of \pi.

    Convert 9i to polars:

    |9i| = 9

    \arg{(9i)} = \frac{\pi}{2}.


    So 9i = 9\,\textrm{cis}\,\frac{\pi}{2}.


    Therefore (9i)^{\frac{1}{2}} = \left(9\,\textrm{cis}\,\frac{\pi}{2}\right)^{\frac  {1}{2}}

     = 9^{\frac{1}{2}}\,\textrm{cis}\,\left(\frac{1}{2}\c  dot \frac{\pi}{2}\right) By DeMoivre's Theorem

     = 3\,\textrm{cis}\,\frac{\pi}{4}.


    And remembering that the other square root only differs by an angle of \pi, that means the other square root is

    3 \,\textrm{cis}\,\frac{5\pi}{4}.



    Converting back to Cartesians:

    3\,\textrm{cis}\,\frac{\pi}{4} = 3\cos{\frac{\pi}{4}} + 3i\sin{\frac{\pi}{4}}

     = \frac{3\sqrt{2}}{2} + i\,\frac{3\sqrt{2}}{2}.


    3\,\textrm{cis}\,\frac{5\pi}{4} = 3\cos{\frac{5\pi}{4}} + 3i\sin{\frac{5\pi}{4}}

     = -\frac{3\sqrt{2}}{2} - i\,\frac{3\sqrt{2}}{2}.
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    Quote Originally Posted by Prove It View Post
    You need to remember that there are exactly two square roots, and they are evenly spaced around the unit circle. In other words, they differ by an angle of \pi.

    Convert 9i to polars:

    |9i| = 9

    \arg{(9i)} = \frac{\pi}{2}.


    So 9i = 9\,\textrm{cis}\,\frac{\pi}{2}.


    Therefore (9i)^{\frac{1}{2}} = \left(9\,\textrm{cis}\,\frac{\pi}{2}\right)^{\frac  {1}{2}}

     = 9^{\frac{1}{2}}\,\textrm{cis}\,\left(\frac{1}{2}\c  dot \frac{\pi}{2}\right) By DeMoivre's Theorem

     = 3\,\textrm{cis}\,\frac{\pi}{4}.


    And remembering that the other square root only differs by an angle of \pi, that means the other square root is

    3 \,\textrm{cis}\,\frac{5\pi}{4}.



    Converting back to Cartesians:

    3\,\textrm{cis}\,\frac{\pi}{4} = 3\cos{\frac{\pi}{4}} + 3i\sin{\frac{\pi}{4}}

     = \frac{3\sqrt{2}}{2} + i\,\frac{3\sqrt{2}}{2}.


    3\,\textrm{cis}\,\frac{5\pi}{4} = 3\cos{\frac{5\pi}{4}} + 3i\sin{\frac{5\pi}{4}}

     = -\frac{3\sqrt{2}}{2} - i\,\frac{3\sqrt{2}}{2}.
    thanks this seems to be reasonable compared to what i done



    but what is the arg(-64i) ??
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    Quote Originally Posted by sandy View Post
    what is the arg(-64i) ??
    If r>0 then \text{Arg}(-ri)=\frac{-\pi}{2}.
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    Quote Originally Posted by sandy View Post
    thanks this seems to be reasonable compared to what i done



    but what is the arg(-64i) ??
    Sorry, but this is absolutley basic. Did you try drawing an Argand diagram and using the definition of arg?
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    Quote Originally Posted by mr fantastic View Post
    Sorry, but this is absolutley basic. Did you try drawing an Argand diagram and using the definition of arg?
    i done the diagram its either -pi/2 or -pi/2 + 2kpi= 3pi/2

    im confused please help
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  10. #10
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    Quote Originally Posted by sandy View Post
    i done the diagram its either -pi/2 or -pi/2 + 2kpi= 3pi/2

    im confused please help
    \arg{z} = -\frac{\pi}{2} + 2\pi k, k \in \mathbf{Z}

    but the Principle Argument

    \textrm{Arg}\,{z} = -\frac{\pi}{2}

    since -\pi < \textrm{Arg}\,{z} \leq \pi.
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  11. #11
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    Quote Originally Posted by Prove It View Post
    \arg{z} = -\frac{\pi}{2} + 2\pi k, k \in \mathbf{Z}

    but the Principle Argument

    \textrm{Arg}\,{z} = -\frac{\pi}{2}

    since -\pi < \textrm{Arg}\,{z} \leq \pi.
    thanks alot so i did use - pi/2 so can you please go and check if the answer is right the one i posted please
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