More Complex Number

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January 25th 2010, 12:10 PM
BabyMilo

More Complex Number

1) Sketch on the Argand diagram the regions which satisfy $1\leq |z+3i| \leq 3$ .

so $1\leq x^2+(y+3)^2 \leq 9$ now what do i do?

2) Indicate clearly on an Argand diagram the region of points that satisfy both the conditions $|z-3+4i|\leq4$ , $|z|\geq |z-10|$

Thanks as always.!
January 25th 2010, 01:39 PM
mr fantastic

Quote:

Originally Posted by BabyMilo

1) Sketch on the Argand diagram the regions which satisfy $1\leq |z+3i| \leq 3$ .

so $1\leq x^2+(y+3)^2 \leq 9$ now what do i do?

2) Indicate clearly on an Argand diagram the region of points that satisfy both the conditions $|z-3+4i|\leq4$ , $|z|\geq |z-10|$

Thanks as always.!

1) $x^2+(y+3)^2 = 1$ and $x^2+(y+3)^2 = 9$ are concentric circles. $1\leq x^2+(y+3)^2 \leq 9$ is the area in between them.

2) $|z-3+4i| \leq 4$ is a circle. $|z| = |z-10|$ is the line x = 5. $|z-3+4i|\leq4$ and $|z|\geq |z-10|$ is the area between the circle and the line (you will need to decide which of the two possible areas it is ....)

By the way, all these relations have a very simple geometric interpretation which makes it easy to get the cartesian equations ....
January 25th 2010, 10:46 PM
BabyMilo

Quote:

Originally Posted by mr fantastic

1) $x^2+(y+3)^2 = 1$ and $x^2+(y+3)^2 = 9$ are concentric circles. $1\leq x^2+(y+3)^2 \leq 9$ is the area in between them.

2) $|z-3+4i| \leq 4$ is a circle. $|z| = |z-10|$ is the line x = 5. $|z-3+4i|\leq4$ and $|z|\geq |z-10|$ is the area between the circle and the line (you will need to decide which of the two possible areas it is ....)

By the way, all these relations have a very simple geometric interpretation which makes it easy to get the cartesian equations ....

I guess i understand what you mean. But the thing is I am very bad at inequalities and geometric. I will post my answers later to check. Thanks!
January 26th 2010, 04:18 AM
HallsofIvy

Do you understand that |z| is the distance from the complex number z to the number 0= 0+ 0i? From that it follows that |a- b| is the distance between the two complex numbers a and b.

Saying that |z+ 3i|= |z- (-3i)|= 1 means that the distance from z to -3i is 1: z can be any number on the circle with center at -3i and radius 1. Saying that $1\le |z+ 3i|$ means that z is on or outside that circle.

Similarly |z+ 3i|= |z- (-3i)|= 4 means that the distance from z to -3i is 4: z can be any number on the circle with center at -3i and radius 4. $|z+ 3i|\le 4$ says that z is on or inside that circle.

Similarly, $|z- 3+ 4i|= |z-(3-4i)|\le 4$ means that z is a point on or inside the circle with center at 3- 4i and radius 4.

$|z|= |z- 10|$ means that the distance from z to the origin is equal to the distance from z to 10. And it is an easy geometry theorem that the set of all point equidistant from two points is the perpendicular bisector of the segement between the two points. Here, that is the line perpendicular to the x-axis (real axis) at (5, 0) which is given by z= 5+ yi for any y.

$|z|\ge |z- 10|$ is that line plus the set of points on the right side of that line so that z is closer to 10 than to 0.
January 26th 2010, 10:12 AM
BabyMilo

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Would this be correct for Q2 ?
thanks!
January 27th 2010, 03:40 AM
mr fantastic

Quote:

Originally Posted by BabyMilo

Would this be correct for Q2 ?
thanks!

The line in your diagram looks nothing like the equation of the line I gave you. And did you read the reply by HallsofIvy who gave the geometric details of how I derived that equation?
January 27th 2010, 03:43 AM
HallsofIvy

Your circle is correct but I don't know how you got that line- and you don't say what line it is. The set of z such that $|z|\ge |z- 10|$ is, as I said, the perpendicular bisector of the line from 0 to 10. The line containing 0 and 10, the real axis, is horizontal, so the perpendicular bisector is the vertical line z= 5+ yi for all y.
January 27th 2010, 06:36 AM
BabyMilo

Quote:

Originally Posted by HallsofIvy

Your circle is correct but I don't know how you got that line- and you don't say what line it is. The set of z such that $|z|\ge |z- 10|$ is, as I said, the perpendicular bisector of the line from 0 to 10. The line containing 0 and 10, the real axis, is horizontal, so the perpendicular bisector is the vertical line z= 5+ yi for all y.

the line is y=5x?
January 27th 2010, 04:58 PM
mr fantastic

Quote:

Originally Posted by BabyMilo

the line is y=5x?

As I said earlier, and which was subsequently explained by HoI, the equation of the line is x = 5. How do you get y = 5x?
January 27th 2010, 10:48 PM
BabyMilo

1 Attachment(s)

Quote:

Originally Posted by mr fantastic

As I said earlier, and which was subsequently explained by HoI, the equation of the line is x = 5. How do you get y = 5x?

would it be like this then?
January 28th 2010, 02:17 AM
mr fantastic

Quote:

Originally Posted by BabyMilo

would it be like this then?

Test a value of z taken from the red area. Does it satisfy the inequality?