Find the set of points z=x+iy from the complex plane which satisfy this conditions:

a)Re(z)\leq0

b)|z|=3


I've solved the first one...

Re(z)=x
so x\leq0

x\in(-infinity,0]

But for the second one? Should I find x, y or should I find z?

Hint: How do you define the modulous of z? It should look like the Pythagorean theorem.

|z|=3=\sqrt{x^2+y^2}

like this?

so

9=x^2+y^2

In this case x,y \in (-infinity,+infinity)??

And what about this:

And what about Re(z)\geq0 , Im(z)\geq0, |z|\leq2 ?

for Re(z), x \in (0,+\propto)

for Im(z) y \in (0,+\propto)

and for |z|\leq2

x^2+y^2=4

What to do next?

z = x+iy
Re(z) = x
Re(z) \leq 0 implies x \leq 0. Job done.

|z| = 3.
z = re^{i\theta} = r\cos \theta + ir\sin \theta = x+iy
What values of r and theta give |z|=3 and what does that mean for x and y?

For the last problem:

Will the soultion be:

x,y \in (0,2) ??

Nope, since z=0 is allowed from that. You've just described an open square centred on the origin. |z|=k is a circle, centred on the origin, with radius k. What's the parametric description for such a curve?

Can somebody please exactly give me what is the set of z points?

Is it x^2+y^2 \leq 4 ?

Is it x
\in (0,+\propto)


Is it

\in (0,+\propto)

Nope, since z=0 is allowed from that. You've just described an open square centred on the origin. |z|=k is a circle, centred on the origin, with radius k. What's the parametric description for such a curve?

|z|=k=\sqrt{(c-x)^2+(d-y)^2}

No, that isn't a parametric description. A parametric description is giving x and y in terms of other variables, which are more convenient.

If |z|=3, then, as I said, it's a circle of radius 3 centred on the origin. Using polar coordinates seems the obvious thing to do so you get that x = r cos(a), y = r sin(a), where r=3 and a \in [0,2\pi).

If |z|<2 then it's x = r cos(a), y = r sin(a), where r<2 and a \in [0,2\pi).

I asked my professor, and he told me that I need to show the results on circular, since the results are:

x^2+y^2\leq4

x \in [0,+\propto)

y \in [0,+\propto)

right?

That cannot be the answer. If x^{2}+y^{2} \leq 4 how can x=y=100 be allowed?

You can quickly see that's wrong by the fact x^{2} \geq 0, so y^{2} \leq 4 so -2 \leq y \leq 2, provided x=0. If x>0 then the allowablew range for y is less.

What about my previous post do you not understand? Surely you see that x=y=100 is wrong, yet you seem to think that ANY non-negative x and y are allowed?!

Can this (http://i27.tinypic.com/30ctztc.jpg) be the solution?

btw- I didn't say that those are the solutions, those are the statements which need to be satisfied.. Sorry for misunderstanding...

That's the region which satisfies |z|<4. So if you're asked for the space which satisfies |z|=4, you'd be wrong.

|z|=4, |z|<4 and |z| \leq 4 are different regions. Summing the first and second regions gives you the third.

Ok, I found that... But now I need help for |z| \leq 4.. Will it be like this? (http://i31.tinypic.com/511rno.jpg)

Please answer me, I need verification. Thank you...

Ok, I solved that. And what about |i|?
Is |i|=\sqrt{i^2}=i?

A modulus is always a real positive number. |x| \not= x^{2}. If z = re^{i\theta} then |z|=r.

So what is "i" equal to? Is it equal to "1"? Why "1"?

You're not equating i to anything. You're computing it's modulus.

|i| = 1 because it's 1 unit away from the origin. |z| is asking "How many units from the origin is z?". i exists in the |z|=1 circle. So you know |i|=1.