Complex Plane Points with Re(z)≤0 and |z|=3

[Redbelly98]

Yes, good job!

You might want to indicate the value of the radius in the figure.

 

[Physicsissuef]

And what about
$$1 \leq |z| \leq 3$$

Will it be, like http://i29.tinypic.com/beabt0.jpg"

 

[Redbelly98]

No.

Try getting the equation (or inequality) in terms of x and y first. By now you've seen enough examples that you should be able to do it.

Then draw the figure.

 

[Physicsissuef]

$$1 \leq x^2+y^2 \leq 9$$
But the question is how will I draw it?

 

[Redbelly98]

Do you honestly not recognize equations for simple familiar figures in what you have written? Even after doing numerous similar problems?

To be honest and frank with you, I am wondering how it will be possible for you to answer questions like this on an exam in the future.

 

[Nick89]

Do you understand that (and why) $$x^2 + y^2 = r^2$$ represents a circle with radius r?

Do you understand that $$x^2 + y^2 < r^2$$ represents a disk with radius r, not including the boundary?

 

[Physicsissuef]

Do you honestly not recognize equations for simple familiar figures in what you have written? Even after doing numerous similar problems?

To be honest and frank with you, I am wondering how it will be possible for you to answer questions like this on an exam in the future.

Redbelly98 please be patient with me... I just started to learn about Complex numbers... I can't be expert with 2 hours of learning for god sake...

Do you understand that (and why) $$x^2 + y^2 = r^2$$ represents a circle with radius r?

Do you understand that $$x^2 + y^2 < r^2$$ represents a disk with radius r, not including the boundary?

For how many times, I understand. $$r^2=x^2+y^2$$ because of Pita gore theorem in polar coordinate system.

Lets go back to the problem. Is it like http://i25.tinypic.com/24wya1d.jpg"

 

[Physicsissuef]

Is it like that, please I need confirmation. Thank you...

 

[dx]

Yes, that's correct.

 

[Physicsissuef]

Ok. Thanks. And for $|z| \leq 4$, is it like http://i31.tinypic.com/511rno.jpg"

 

[dx]

Yes.

 

[Physicsissuef]

Ok, thanks. And what about this http://i26.tinypic.com/jj360y.jpg"

a)$$1 \leq |z| \leq 2$$

b)

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


$$-2 \leq Im(z) \leq 2$$

c)$$|z-(1,1)| \leq \sqrt{2}$$

?

 

[dx]

(a) and (c) are correct, (b) is wrong.

 

[Physicsissuef]

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

$$0 \leq Im(z) \leq 2$$

It was typo. Now is it correct?

 

[dx]

Yes.

 

[Physicsissuef]

Ok, thanks. And can you please tell me about this two pictures. http://i31.tinypic.com/18d3py.jpg" I will be very happy if you can confirm me. Thank you once again...

 

[HallsofIvy]

Ok. Thanks. And for $|z| \leq 4$, is it like http://i31.tinypic.com/511rno.jpg"

Yes, $|z|\le 4$, in the complex plane, refers to those points, (x,y), whose distance from (0,0) is less than or equal to 4. That consists of the circle with center at (0,0) and radius 4 as well as its interior.

 

[Physicsissuef]

Yes, but can you confirm me about the pictures in the above post, they're not same. Thanks.

 

[dx]

Ok, thanks. And can you please tell me about this two pictures. http://i31.tinypic.com/18d3py.jpg" I will be very happy if you can confirm me. Thank you once again...

Yes they're correct.

 

[Physicsissuef]

They are all correct? What about i) $0 < |z| \leq 3$? How will I make the disk without zero?

 

[Nick89]

Usually you put a circle (like an 'o') on the point, instead of a dot.

For example, this could be the line $y = 2x + 1, x \neq 3$:

|    /
|   / 
|  o
| /
|/
|_________

 

[Physicsissuef]

Ok, thank you, and my other tasks are correct? (I mean, with stuff like shadowing and circles like in this case) Is it correct with the angles?

 

[Nick89]

Yes. Don't worry about these things (how to draw it) too much, as long as it's clear what you mean. If your teacher understands what you mean he will most probably accept it, even if the way you have drawn it is maybe not 100% 'conventional'... (I don't even know if there is any convention for things like this... But it's pretty obvious how you've drawn it, so I think you're going to be ok.)

 

[Physicsissuef]

Ok. Thank you.