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Math Help - sketch

  1. #1
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    sketch

    Sketch Re(z^4). ie:sketch real part of z^4
    I tried to do the arithmetic and I think Re(z^4)=x^4 - 6x^2*y^2 + y^4
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  2. #2
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    Quote Originally Posted by scubasteve123 View Post
    Re(z^4)=x^4 - 6x^2*y^2 + y^4
    I agree with this.
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  3. #3
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    Quote Originally Posted by scubasteve123 View Post
    Sketch Re(z^4). ie:sketch real part of z^4
    I tried to do the arithmetic and I think Re(z^4)=x^4 - 6x^2*y^2 + y^4
    z^4 = (x + iy)^4

     = \sum_{k = 0}^4 {4\choose{k}}x^{4 - k}(iy)^k

     = x^4 + 4ix^3y + 6i^2x^2y^2 + 4i^3xy^3 + i^4y^4

     = x^4 + 4ix^3y - 6x^2y^2 - 4ixy^3 + y^4

     = x^4 - 6x^2y^2 + y^4 + i(4x^3y - 4xy^3).


    So \mathbf{Re}(z) = x^4 - 6x^2y^2 + y^4


    You are correct with your calculation.

    Are you given any more information about z? Because you will need to evaluate \mathbf{Re}(z) = \textrm{constant} in order to make the sketch...
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  4. #4
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    Quote Originally Posted by scubasteve123 View Post
    Sketch Re(z^4). ie:sketch real part of z^4
    I tried to do the arithmetic and I think Re(z^4)=x^4 - 6x^2*y^2 + y^4
    Your expression for Re(z^4) is correct. But there is no equation and therefore nothing to sketch unless Re(z^4) is equal to something ....
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Your expression for Re(z^4) is correct. But there is no equation and therefore nothing to sketch unless Re(z^4) is equal to something ....
    Hi everyone thanks for checking I was so conserned with my arithmetic i forgot the rest my apologies. I need to sketch {z element complex numbers such that Re(z^4)>0 and |z-1|<2}
    I know how to sketch the |z-1|<2 part and i assume the part I want is where the 2 overlap i just not sure how to sketch x^4-6x^2*y^2 +y^4
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  6. #6
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    Im new to sketching in complex.
    should i be trying to factor the expression I found for Re(z^4) ? or should i be able to sketch just using my expression? How do i know what this shape consists of? also since its only the real part is it just a line?
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  7. #7
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    Solve x^4 - 6x^2y^2 + y^4 > 0 for y.

    I would advise completing the square...

    y^4 - 6x^2y^2 + x^4 > 0

    y^4 - 6x^2y^2 + (-3x^2)^2 - (-3x^2)^2 + x^4 > 0

    (y^2 - 3x^2)^2 - 8x^4 > 0

    (y^2 - 3x^2)^2 > 8x^4

    |y^2 - 3x^2| > 2\sqrt{2}x^2

    y^2 - 3x^2 < -2\sqrt{2}x^2 or y^2 - 3x^2 > 2\sqrt{2}x^2


    Can you go from here?
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  8. #8
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    would i now combine the x^2 terms then i get y^2>(2*2^1/2 +3)x^2
    and y^2<(-2*2^1/2 +3)x^2 so now i see that the graph is greater than 2*2^1/2 +3 and less than -2*2^1/2 +3 ... im confused.
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  9. #9
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    Case 1:

    y^2 < (3 - 2\sqrt{2})x^2

    |y| < x\sqrt{3 - 2\sqrt{2}}

    x\sqrt{2\sqrt{2} - 3} < y < x\sqrt{3 - 2\sqrt{2}}


    Case 2:

    y^2 - 3x^2 > 2\sqrt{2}x^2

    y^2 > (3 + 2\sqrt{2})x^2

    |y| > \sqrt{(3 + 2\sqrt{2})x^2}

    y <x\sqrt{- 3 - 2\sqrt{2}} or y > x\sqrt{3 + 2\sqrt{2}}.

    Now graph these inequalities.
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  10. #10
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    Quote Originally Posted by Prove It View Post
    Case 1:

    y^2 < (3 - 2\sqrt{2})x^2

    |y| < x\sqrt{3 - 2\sqrt{2}}

    x\sqrt{2\sqrt{2} - 3} < y < x\sqrt{3 - 2\sqrt{2}}


    Case 2:

    y^2 - 3x^2 > 2\sqrt{2}x^2

    y^2 > (3 + 2\sqrt{2})x^2

    |y| > \sqrt{(3 + 2\sqrt{2})x^2}

    y <x\sqrt{- 3 - 2\sqrt{2}} or y > x\sqrt{3 + 2\sqrt{2}}.

    Now graph these inequalities.

    Im hoping that im finally having an "eureka" moment.. okay thank you for the help finding the inequalities... now to graph the inequalities do i just graph them like i would normally on the x,y plane since its only the Real part of the complex number??
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  11. #11
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    Quote Originally Posted by scubasteve123 View Post
    Im hoping that im finally having an "eureka" moment.. okay thank you for the help finding the inequalities... now to graph the inequalities do i just graph them like i would normally on the x,y plane since its only the Real part of the complex number??
    Yes graph them as you normally would on the x, y plane.

    Just name your axes as \mathbf{Re}(z) and \mathbf{Im}(z), since x = \mathbf{Re}(z) and y = \mathbf{Im}(z).
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  12. #12
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    u've been so helpful. its really appreciated
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