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Thread: complex numbers

  1. #1
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    complex numbers

    complex numbers-win_20170905_10_55_50_pro.jpg

    i am stuck on question 12 I worked out part (ii) by saying z^3 = x(cos360 +isin360) therefore z=x(cos120+isin120). i am not sure how to do part (iii)
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  2. #2
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    Re: complex numbers

    Quote Originally Posted by edwardkiely View Post
    Click image for larger version. 

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    i am stuck on question 12 I worked out part (ii) by saying z^3 = x(cos360 +isin360) therefore z=x(cos120+isin120).
    This is incorrect. It should be z= x^{1/3}W(cos(120)+ i sin(120)).

    If z= x(cos(\theta)+ i sin(\theta)) then |z|= x so [tex]|^3|= (9/4)|z| then [tex]x^3= (9/4)x. 4x^3- 9x= x(4x^2- 9)= 0. Can you solve for x?

    i am not sure how to do part (iii)
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  3. #3
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    Re: complex numbers

    what does W stand for?

    i am sorry but i can not understand you last line.
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  4. #4
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    Re: complex numbers

    Quote Originally Posted by edwardkiely View Post
    what does W stand for?
    i am sorry but i can not understand you last line.
    That W is just a miss-type of a random symbol.

    Here is what is going on:
    $ \begin{align*}\text{Because }|z^3|&=|z|^3 \\|z|^3&=\dfrac{9}{4}|z|\\|z|^2&=\dfrac{9}{4}\\|z| &=\dfrac{\pm 3}{2} \end{align*}$

    Can you now finish?
    Thanks from edwardkiely
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  5. #5
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    Re: complex numbers

    Quote Originally Posted by HallsofIvy View Post
    This is incorrect. It should be z= x^{1/3}W(cos(120)+ i sin(120)).

    If z= x(cos(\theta)+ i sin(\theta)) then |z|= x so [tex]|^3|= (9/4)|z| then [tex]x^3= (9/4)x. 4x^3- 9x= x(4x^2- 9)= 0. Can you solve for x?
    Thanks, Plato. Yes, the "W" was a mistype. It should have been z= x^{1/3}(cos(120)+ i sin(120)). The point is that, with |z|= x, you should have had " x^{1/3}" not " x".

    x^3= (9/4)x is equivalent to 4x^3- 9x= x(4x^2- 9)= 0 so either x= 0 or 4x^2- 9= (2x- 3)(2x+ 3)= 0.

    Of course, in order to have |z|= x, with z non-zero as shown, x must be a positive real number.
    Last edited by HallsofIvy; Sep 5th 2017 at 09:05 AM.
    Thanks from edwardkiely
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  6. #6
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    Re: complex numbers

    thanks
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