# 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?

Thanks in advance,
Dan.

2. Oct 10, 2007

### mjsd

from description, it appears that this is your task:
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

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