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Math Help - [SOLVED] Complex analysis, determine the set in the complex plane that satisfies...

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    MHF Contributor arbolis's Avatar
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    [SOLVED] Complex analysis, determine the set in the complex plane that satisfies...

    I've been stuck for 3 days on this simple problem. I must graph the set of the complex plane where z satisfies |3z+2|<1. I've used the way of writing z=re^{i\theta} but reached a non sense condition for r. Well, the equation |3z+1|=1 would hold if r=-\frac{1}{3 \cos \theta} \pm \frac{\sqrt {-2-\sin ^2 \theta}}{3 \cos \theta}. I really have no idea where I went wrong. I know there's likely an easy way to solve the problem, but still, I don't know where I'm wrong by replacing z for re^{i\theta} and then calculate the modulus of 3z+2, etc.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by arbolis View Post
    I've been stuck for 3 days on this simple problem. I must graph the set of the complex plane where z satisfies |3z+2|<1. I've used the way of writing z=re^{i\theta} but reached a non sense condition for r. Well, the equation |3z+1|=1 would hold if r=-\frac{1}{3 \cos \theta} \pm \frac{\sqrt {-2-\sin ^2 \theta}}{3 \cos \theta}. I really have no idea where I went wrong. I know there's likely an easy way to solve the problem, but still, I don't know where I'm wrong by replacing z for re^{i\theta} and then calculate the modulus of 3z+2, etc.
    In general, \left\{z\in\mathbb{C}:|az+b|<r\right\} is a disk of radius r centered around...what?
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by Drexel28 View Post
    In general, \left\{z\in\mathbb{C}:|az+b|<r\right\} is a disk of radius r centered around...what?
    My bad, a semester of physics totally destroyed me...
    Around -b?
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    Quote Originally Posted by arbolis View Post
    I've been stuck for 3 days on this simple problem. I must graph the set of the complex plane where z satisfies |3z+2|<1. I've used the way of writing z=re^{i\theta} but reached a non sense condition for r. Well, the equation |3z+1|=1 would hold if r=-\frac{1}{3 \cos \theta} \pm \frac{\sqrt {-2-\sin ^2 \theta}}{3 \cos \theta}. I really have no idea where I went wrong. I know there's likely an easy way to solve the problem, but still, I don't know where I'm wrong by replacing z for re^{i\theta} and then calculate the modulus of 3z+2, etc.

    Hint - Screw the polar form: write z=x+iy\,,\,\,x,y\in\mathbb{R}, substitute in |3x+2|<1 , square and do a little algebra there.

    You must get the set of all points inside a certain circle (with center on the negative x-axis) of radius \frac{1}{3} ...

    A big deal of problems of this kind can easily be solved as above.

    Tonio
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by arbolis View Post
    My bad, a semester of physics totally destroyed me...
    Around -b?
    Note that literally |z-z_0|<r is the open disk of radius r around z_0. So, if a\ne 0 then |az+b|<r\Leftrightarrow |z-\tfrac{-b}{a}|<\frac{r}{|a|}
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    MHF Contributor arbolis's Avatar
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    I had tried at first the way of writing z=x+iy but reached non sense probably due to arithmetic errors. Like now: I reached \sqrt {9x^2 +6x +4 + 9 y^2}<1. So 9x^2 +6x +4 + 9 y^2<1. Skipping the steps (with 1 completion of square) I reach \left ( x + \frac{1}{3} \right ) ^2 + y^2 < -\frac{2}{9}. So 2 positive numbers add to a negative one... Are the 2 first inequations right at least?
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    Quote Originally Posted by arbolis View Post
    I had tried at first the way of writing z=x+iy but reached non sense probably due to arithmetic errors. Like now: I reached \sqrt {9x^2 +6x +4 + 9 y^2}<1. So 9x^2 +6x +4 + 9 y^2<1. Skipping the steps (with 1 completion of square) I reach \left ( x + \frac{1}{3} \right ) ^2 + y^2 < -\frac{2}{9}. So 2 positive numbers add to a negative one... Are the 2 first inequations right at least?

    No, they aren't. Why is it that sometimes people rushes to open up parentheses? Leave that as a last resource!

    |3z+2|<1\iff |3x+2+3yi|^2<1\iff (3x+2)^2+9y^2<1 \iff 9\left(x+2/3\right)^2+9y^2<1\iff  \left(x+2/3\right)^2+y^2<\frac{1}{9} ...

    Drexel's solution though is more elegant and short, once you know that |z-z_0|<a\,,\,0<a\in\mathbb{R} , is the inside of a circle.

    Tonio
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    In general, \left\{z\in\mathbb{C}:|az+b|<r\right\} is a disk of radius r centered around...what?
    (Of radius r/|a|)
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    (Of radius r/|a|)
    Typo, clearly fixed in my next post.
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    The point people are trying to make to you is to do this "geometrically" not "algebraically".

    In the complex numbers, |z- a| is the distance between z and a. |z- a|= r is satisfied by all z on the circle with center a and radius r. |z- a|< r is the open disk of point inside that circle.

    Here, your problem was |3z+ 2|< 1 which is the same as 3|z+ 2/3|< 1 or
    |z-(-2/3)|< 1/3.
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    MHF Contributor arbolis's Avatar
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    Ok thanks to all, I get it. So the center of the disk is at -2/3, right?

    I'm still totally puzzled why (3x+2)^2+9y^2<1 is true while 9x^2 +6x +4 + 9 y^2<1 is not true. I know I don't need to solve the problem this way, but still, I'm stuck at understanding why it isn't true.

    Edit: Oops, ok I get it. I forgot a factor 2. Problem solved.
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    Quote Originally Posted by arbolis View Post
    Ok thanks to all, I get it. So the center of the disk is at -2/3, right?

    I'm still totally puzzled why (3x+2)^2+9y^2<1 is true while 9x^2 +6x +4 + 9 y^2<1 is not true. I know I don't need to solve the problem this way, but still, I'm stuck at understanding why it isn't true.

    Edit: Oops, ok I get it. I forgot a factor 2. Problem solved.

    Indee, and the center is a point on the plane, not a number: (-2/3,\,0)

    Tonio
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    MHF Contributor arbolis's Avatar
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    Quote Originally Posted by tonio View Post
    Indee, and the center is a point on the plane, not a number: (-2/3,\,0)

    Tonio
    Yes sorry for being imprecise. I meant the point on the real axis.
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