I've been stuck for 3 days on this simple problem. I must graph the set of the complex plane where z satisfies . I've used the way of writing but reached a non sense condition for r. Well, the equation |3z+1|=1 would hold if . I really have no idea where I went wrong. I know there's likely an easy way to solve the problem, but still, I don't know where I'm wrong by replacing z for and then calculate the modulus of , etc.
Hint - Screw the polar form: write , substitute in , square and do a little algebra there.
You must get the set of all points inside a certain circle (with center on the negative x-axis) of radius ...
A big deal of problems of this kind can easily be solved as above.
Tonio
I had tried at first the way of writing z=x+iy but reached non sense probably due to arithmetic errors. Like now: I reached . So . Skipping the steps (with 1 completion of square) I reach . So 2 positive numbers add to a negative one... Are the 2 first inequations right at least?
The point people are trying to make to you is to do this "geometrically" not "algebraically".
In the complex numbers, |z- a| is the distance between z and a. |z- a|= r is satisfied by all z on the circle with center a and radius r. |z- a|< r is the open disk of point inside that circle.
Here, your problem was |3z+ 2|< 1 which is the same as 3|z+ 2/3|< 1 or
|z-(-2/3)|< 1/3.
Ok thanks to all, I get it. So the center of the disk is at -2/3, right?
I'm still totally puzzled why is true while is not true. I know I don't need to solve the problem this way, but still, I'm stuck at understanding why it isn't true.
Edit: Oops, ok I get it. I forgot a factor 2. Problem solved.