1. The problem statement, all variables and given/known data

What will be solutions of this (http://i26.tinypic.com/jj360y.jpg) complex circles.


2. Relevant equations

z=a+bi

3. The attempt at a solution


a)
1 \leq Re(z) \leq 2

1 \leq Im(z) \leq 2

What about z?

b)
-2 \leq Re(z) \leq 2

-2 \leq Im(z) \leq 2

What about z?

c) I don't know. I think |z|=\sqrt{2}

a)
1 \leq Re(z) \leq 2

1 \leq Im(z) \leq 2
When you see circles ALWAYS work in polar coordinates. Because if you work in cartesians you end up with rectangles. For instance, your attempt says that I'm allowed z = 1.9+1.9i, which isn't in the region shaded. Just draw the region you just suggested. It's the square with corners at z=2+2i, 2-2i, -2+2i, -2-2i minus the square with corners at z=1+i, 1-i, -1+i, -1-i.

Your relevent equations are useless. A more relevent equation would be something like :

The equation of a circle centrated at z=a+ib with radius R is |z-(a+ib)| = R

Can you see why that's true? And much more useful?

It instantly tells you your answer to c) is wrong. |z| = sqrt(2) is a circle centred AT THE ORIGIN of radius sqrt(2). You claim -1-i is in the region. Is it? Check your 'suggested answers'. Because it's clear you are not and they are almost always easily shown to be wrong, unfortunately.

a) 1 \leq |z| \leq 2
Is it better now? Should I find set for Re(z) and Im(z) ??

Please be patient... I just started this Topic for Complex numbers... Thanks...

Is it correct for a)?

And is for b)

http://www.sciforums.com/cgi-bin/mimetex.cgi?-2%20\leq%20Re(z)%20\leq%202
http://www.sciforums.com/cgi-bin/mimetex.cgi?-2%20\leq%20Im(z)%20\leq%202