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

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

AND

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 or sth else)?

Originally Posted by courteous

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

AND

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 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 . 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 . You want the outside of this circle and its boundary.

To do it algebraically, I'll return later ....

Originally Posted by mr fantastic

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 . 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 . 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:

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