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Math Help - polar form, urgent

  1. #1
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    polar form, urgent

    ok, exam in 2 hours and i dont need the answer to this question, i need a list of the steps i take in order to do it, would be hugely grateful.

    z = 1 - i
    w = 1+ i

    a) write in polar form
    b) evaluate z^3/w^7 in form a+ib
    c) find all cube roots of z leaving answers in polar form

    thanks if anyone can help
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    don't you think this is cutting it a bit close. asking for help 2 hours before an exam?
    Quote Originally Posted by Revolt View Post
    ok, exam in 2 hours and i dont need the answer to this question, i need a list of the steps i take in order to do it, would be hugely grateful.

    z = 1 - i
    w = 1+ i

    a) write in polar form
    the polar form of a complex number a + ib is: re^{i \theta}

    where r = \sqrt{a^2 + b^2} and \theta = \arctan \frac ba (usually)


    b) evaluate z^3/w^7 in form a+ib
    find and simplify \frac {z^3}{w^7} with z and w in their polar forms. then convert back to a + ib form


    c) find all cube roots of z leaving answers in polar form
    see post #4 here

    of course, we would write e^{i \theta} instead of \cos \theta + i \sin \theta as was done there

    try them and tell me what you get. it will be good practice for that exam of yours
    Last edited by Jhevon; January 13th 2008 at 10:37 PM.
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  3. #3
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    Quote Originally Posted by Revolt View Post
    ok, exam in 2 hours and i dont need the answer to this question, i need a list of the steps i take in order to do it, would be hugely grateful.

    z = 1 - i
    w = 1+ i

    a) write in polar form
    b) evaluate z^3/w^7 in form a+ib
    c) find all cube roots of z leaving answers in polar form

    thanks if anyone can help
    (a)

    z=r e^{i \theta}=r \cos(\theta) + r \sin(\theta) = 1-i

    so:

    r=\sqrt{2} ,

    and \theta is the angle such that:

    \cos(\theta)=1/\sqrt(2)

    and

    \sin(\theta)=-1/\sqrt(2).


    A similar procedure will turn w into polar form: w=s e^{i \phi}

    (b) z^3/w^7 = (r^4/s^7)(e^{i 3 \theta}/e^{i 7 \phi})= (r^4/s^7)e^{i (3 \theta-7 \phi)}

    which you should be able to turn back into cartesian form.

    (c) z=r e^{i \theta} where r and \theta take the values found earlier.

    Then we may write:

    z=r e^{i \theta+2 i n \pi},\ n\in \mathbb{Z} (an integer)

    then

    z^{1/3}=r^{1/3} e^{(i \theta+2 i n \pi)/3},\ n\in \mathbb{Z}

    and there are just three distinct values for this corresponding to n=0,1,2

    RonL
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  4. #4
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    z = 1 - i
    w = 1+ i

    a) write in polar form
    Convert to polar form:

    find the modulus |z|(also written as r), which is \sqrt{(x^2 + y^2)}
    in the case of z \sqrt{(1^2 + 1^2)}= \sqrt 2
    z = r = \sqrt2
    Then find the argument  \theta . Since tan  \theta = \frac {rise}{run} you can easily find the argument here.

    Polar form would be r (cos \theta + isin \theta)

    b) evaluate z^3/w^7 in form a+ib
    To find exponents of complex numbers, use De Moivre's theorem

    If  z = r (cos \theta + isin \theta)
    then  z^n = r^n (cos (n\theta) + isin (n\theta))

    After this, convert back to a+ib form by multiplying r to (cos \theta + isin \theta)

    c) find all cube roots of z leaving answers in polar form
    Assume  z = m^3 where m is a cube root of z

    so  m = \sqrt[3]r (cos \frac {\theta}{3} + i sin \frac {\theta}{3})
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  5. #5
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    Quote Originally Posted by Jhevon View Post

    where r = \sqrt{a^2 + b^2} and \theta = \arctan \frac ba (usually)
    \arctan does not do here as we need the angle in the range [0 , 2 \pi) (or equivalent) but \arctan gives an answer in the range (\pi/2,\pi/2] (or equivalent).

    RonL
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  6. #6
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    Quote Originally Posted by noreaction View Post
    c) find all cube roots of z leaving answers in polar form

    Assume  z = m^3 where m is a cube root of z

    so  m = \sqrt[3]r (cos \frac {\theta}{3} + i sin \frac {\theta}{3})
    How many cube roots of z does that give us? Should there not be 3?

    RonL
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  7. #7
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    Sorry, i forgot to put +2k\pi for the angles
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