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Math Help - sketch on complex plane

  1. #1
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    sketch on complex plane

    sketch abs(3z-2+5i)<3

    i replace z with rcis(theta)

    then do you move the other terms the the other side of the inequality and use them as asymptotes?

    so it would become a circle radius 3r and asymptotes at 5 and -5i?
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  2. #2
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    Quote Originally Posted by CarmineCortez View Post
    sketch abs(3z-2+5i)<3

    i replace z with rcis(theta)

    then do you move the other terms the the other side of the inequality and use them as asymptotes?

    so it would become a circle radius 3r and asymptotes at 5 and -5i?
    |3z - 2 + 5i| < 3 \Rightarrow 3 \left|z - \left(\frac{2}{3} - \frac{5}{3} i \right) \right| < 3 \Rightarrow \left|z - \left(\frac{2}{3} - \frac{5}{3} i \right) \right| < 1.

    The boundary of the required region is a circle of radius 1 and centre at z = \frac{2}{3} - \frac{5}{3} i (geometrically, do you know why?)

    Now you have to determine whether the required region is inside or outside this circle.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    |3z - 2 + 5i| < 3 \Rightarrow 3 \left|z - \left(\frac{2}{3} - \frac{5}{3} i \right) \right| < 3 \Rightarrow \left|z - \left(\frac{2}{3} - \frac{5}{3} i \right) \right| < 1.

    The boundary of the required region is a circle of radius 1 and centre at z = \frac{2}{3} - \frac{5}{3} i (geometrically, do you know why?)

    Now you have to determine whether the required region is inside or outside this circle.

    the region should be inside the circle since we have abs(z-...)<1

    so the goal is to have no coefficient for z? or a 1 for the inequality. I'm a little confused here.
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  4. #4
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    Sketch on complex plane

    Hello CarmineCortez
    Quote Originally Posted by CarmineCortez View Post
    the region should be inside the circle since we have abs(z-...)<1

    so the goal is to have no coefficient for z? or a 1 for the inequality. I'm a little confused here.
    The goal is to have 'no' coefficient for z - although, of course, the coefficient is really 1. (Don't forget that, will you?)

    The reason this works is that geometrically (to use Mr F's word) |z - a| is the distance between the points representing the complex numbers z and a. So if a is a constant, |z - a| will be the distance of the point representing z from a fixed point, which we can call A. This in turn is why the equation |z - a| = r (where r is real) represents a circle, centre A, radius r.

    OK now?

    Grandad
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