Complex Plane

1. Oct 10, 2007

danago

Given that $$|z-1+i| \le 1$$, find the maximum and minimum value of |z| and Arg(z).

I realise that the equation given defines the interior of a circle of radius 1 centered at (1,-1), which includes the circumference.

For the first part of the question, i am able to represent the equation graphically. From what i understand, |z| is the distance from the origin to any point lying on or within the circle. If this is the case, i can see the minimum and maximum points, but im not too sure on how to calculate their locations.

For the next part, finding the extreme values of Arg(z), i just read straight from my graph and said that the minimum is $$-\pi/2$$ and the maximum is 0. Is that right?

Dan.

2. Oct 10, 2007

mjsd

given the set of points defined by $$\{z:|z-1+i|\leq 1\}$$ find the complex numbers z such that the vector going from origin to z has max/min length. likewise for angle (so -p1/2 and 0 seem wrong). but then again you said you read straight from your graph, how does your graph look like, or how you derived it? (to help with pin-pointing potential mistakes)

EDIT: sorry my mistake

Last edited: Oct 10, 2007
3. Oct 10, 2007

danago

I simply drew a circle centered at (1,-1) with radius 1. Both the x and y axis are tangental to the circle. That pretty much explains what ive drawn.

My book says the argument of a complex number should be defined within -pi to pi, which is how i got 0 and -pi/2, since the circle touches the positive x axis and the negative y axis.

4. Oct 10, 2007

Dick

Ok, seems like your picture is fine. For |z|, how far is the center of the circle from the origin? Now how far are the closest and farthest points from the center of the circle?

5. Oct 10, 2007

danago

Ah ok thats a good way to think about it. The distance from the origin to (1,-1) is $$\sqrt{2}$$, plus another 1 unit (the circles radius) gives a maximum distance of $$\sqrt{2}+1$$. The minimum distance will just then be $$\sqrt{2}-1$$. Am i right?

6. Oct 10, 2007

Dick

Absolutely.

7. Oct 10, 2007

danago

Alright thanks for the help What about the argument of z? Was i right with that?

8. Oct 10, 2007

HallsofIvy

Staff Emeritus
Yes, you were.

9. Oct 10, 2007

danago

Alright, thanks alot