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Math Help - Trying to sketch this graph on the complex plane

  1. #1
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    Trying to sketch this graph on the complex plane

    Sketch the graph of |z+5| = 4; that is, find all z \in \mathbb{C} which satisfy this equation. What geometric shape is it?

    |x + iy +5| = 4

    |(x+5) +iy|=4

    \sqrt{(x+5)^2+y^2} = 4

    (x+5)^2+y^2=16

    x^2+10x+25 +y^2 = 16

    x^2 + y^2 = -10x - 9

    I don't know what to do here. Is it a circle?

    My notes have steps in polar form:

    (\sqrt3)^4(cos\left(\frac{8\pi}{3}\right)+isin \left( \frac{8\pi}{3}\right) )

    ...

    = 9(-\frac{1}{2}+i\frac{\sqrt3}{2})

    I'm not sure how these were derived.
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  2. #2
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    Re: Trying to sketch this graph on the complex plane

    It is a circle with radius 4 and centre at (x,y) = (-5,0).
    Last edited by sander; September 10th 2011 at 11:06 PM. Reason: removed incorrect remark
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  3. #3
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    Re: Trying to sketch this graph on the complex plane

    Quote Originally Posted by sander View Post
    Btw. I think your third, fourth, fifith and six line should include -y^2 instead of y^2.
    Why? I didn't square i.
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  4. #4
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    Re: Trying to sketch this graph on the complex plane

    because y is the imaginary part, not iy, therefore you dont square i along with y.

    it is a circle with radius 4 and center at (-5,0) as Sander pointed out
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  5. #5
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    Re: Trying to sketch this graph on the complex plane

    Sander and andrew2322 are using the fact that |z- a| is the distance from the point z to the point a in the complex plane. |z+ 5|= |z- (-5)| so |z+ 5|= 4 is satisfied for all points whose distanced from -5 is equal to 4- in other words, the circle with center -5 and radius 4.

    Also, you arrived at (x+ 5)^2+  y^2=  16 and then proceeded to multiply out the square, etc. You should not have done that. You should have recognized immediately that this is of the form (x- a)^2+ (y- b)^2= r^2, the equation of a circle with center (a, b) and radius r, with a= -5, b= 0, and r= 4.
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  6. #6
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    Re: Trying to sketch this graph on the complex plane

    Quote Originally Posted by HallsofIvy View Post
    Sander and andrew2322 are using the fact that |z- a| is the distance from the point z to the point a in the complex plane. |z+ 5|= |z- (-5)| so |z+ 5|= 4 is satisfied for all points whose distanced from -5 is equal to 4- in other words, the circle with center -5 and radius 4.

    Also, you arrived at (x+ 5)^2+  y^2=  16 and then proceeded to multiply out the square, etc. You should not have done that. You should have recognized immediately that this is of the form (x- a)^2+ (y- b)^2= r^2, the equation of a circle with center (a, b) and radius r, with a= -5, b= 0, and r= 4.
    That made a lot of sense, thanks! Yah, I didn't realise it was a circle. I only knew it in the form x^2 + y^2 = r^2
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