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Thread: 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 $\displaystyle a + ib$ is: $\displaystyle re^{i \theta}$

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


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


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

    of course, we would write $\displaystyle e^{i \theta}$ instead of $\displaystyle \cos \theta + i \sin \theta$ 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; Jan 13th 2008 at 09:37 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)

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

    so:

    $\displaystyle r=\sqrt{2}$ ,

    and $\displaystyle \theta$ is the angle such that:

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

    and

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


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

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

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

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

    Then we may write:

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

    then

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

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

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

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

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

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

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

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

    so $\displaystyle 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 $\displaystyle r = \sqrt{a^2 + b^2}$ and $\displaystyle \theta = \arctan \frac ba$ (usually)
    $\displaystyle \arctan$ does not do here as we need the angle in the range $\displaystyle [0 , 2 \pi)$ (or equivalent) but $\displaystyle \arctan$ gives an answer in the range $\displaystyle (\pi/2,\pi/2]$ (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 $\displaystyle z = m^3$ where m is a cube root of z

    so $\displaystyle m = \sqrt[3]r (cos \frac {\theta}{3} + i sin \frac {\theta}{3}) $
    How many cube roots of $\displaystyle z$ 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 $\displaystyle +2k\pi $ for the angles
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