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Math Help - |z-3|<=5 ^ |z+i|>=1 (complex numbers)

  1. #1
    Member courteous's Avatar
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    |z-3|<=5 ^ |z+i|>=1 (complex numbers)

    First, hello to everyone. (Did I put my thread in the right category?)

    |z-3|<=5 AND |z+i|>=1

    You must draw a graph from both conditions (both are curves). I can easily visualize what the outcome would be (with the help of curve equation). For instance, |z+i|>=1, obviously, has its curve centered at (0, -i) and has radius of more than 1.

    But how do you mathematically derive this (using z=a+bi or sth else)?
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  2. #2
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    Quote Originally Posted by courteous View Post
    First, hello to everyone. (Did I put my thread in the right category?)

    |z-3|<=5 AND |z+i|>=1

    You must draw a graph from both conditions (both are curves). I can easily visualize what the outcome would be (with the help of curve equation). For instance, |z+i|>=1, obviously, has its curve centered at (0, -i) and has radius of more than 1.

    But how do you mathematically derive this (using z=a+bi or sth else)?
    Geometrically, |z - 3| = 5 gives all the points a distance of 5 from z = 3. This is a circle with radius 5 and centre at z = 3, that is, (3, 0). So the Cartesian equation is (x - 3)^2 + y^2 = 5^2. You want the inside of this circle and its boundary.

    Similarly, |z + i| = 1 <=> |z - (-i)| = 1 gives all the points a distance of 1 from z = -i. This is a circle with radius 5 and centre at z = -i, that is, (0, -1). So the Cartesian equation is x^2 + (y + 1)^2 = 1^2. You want the outside of this circle and its boundary.

    To do it algebraically, I'll return later ....
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Geometrically, |z - 3| = 5 gives all the points a distance of 5 from z = 3. This is a circle with radius 5 and centre at z = 3, that is, (3, 0). So the Cartesian equation is (x - 3)^2 + y^2 = 5^2. You want the inside of this circle and its boundary.

    Similarly, |z + i| = 1 <=> |z - (-i)| = 1 gives all the points a distance of 1 from z = -i. This is a circle with radius 5 and centre at z = -i, that is, (0, -1). So the Cartesian equation is x^2 + (y + 1)^2 = 1^2. You want the outside of this circle and its boundary.

    To do it algebraically, I'll return later ....
    Consider |z + i| = 1 and substitute z = x + iy:

    |x + iy + i| = 1 \Rightarrow |x + i(y+1)| = 1 \Rightarrow \sqrt{x^2 + (y+1)^2} = 1 \Rightarrow x^2 + (y+1)^2 = 1.
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