Sketching the closure of a set

In summary, the homework statement is that Re(1/z) = Re(x-iy)/(x^2 + y^2), only if x less than or equal to 1/2 and y is less than or equal to zero. 1/2 lies only the real part of the complex plane since 1/2=1/2+i*0.
  • #1
Benzoate
422
0

Homework Statement



Sketch the closure of the set:Re(1/z)=< 1/2[ b]2. Homework Equations [/b]

The Attempt at a Solution



Re(1/z)=Re(1/(x+iy)

Re(1/(x+iy))=< 1/2. Not really sure how to sketch 1/2 on a complex plane. Maybe 1/2 can be written in a complex form: 1/2= (1/2)+(0)*i=1/2 and therefore , 1/2 only lies exclusively on the x-axis.

Not even entirely sure how to sketch the Re(1/z) on a complex plane. Generally, Re(z)=x and therefore would Re(1/z)=1/x?
 
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  • #2
Benzoate said:
Re(1/z)=Re(1/(x+iy)

Hi Benzoate! :smile:

ok … now you need to simplify this fraction so that you can just read off the real part.

The standard trick is to multiply top and bottom by (x - iy). :smile:
 
  • #3
tiny-tim said:
Hi Benzoate! :smile:

ok … now you need to simplify this fraction so that you can just read off the real part.

The standard trick is to multiply top and bottom by (x - iy). :smile:

so Re (1/z) = Re(1/(x+iy))=Re(1/(x+iy)*(x-iy)/(x-iy))=Re(x-iy)/(x^2-y^2)). How will this new expression make it easier for me to mapped Re(1/z) in a complex plane. In addition, how would I mapped 1/2 on a complex plane? Does 1/2=1/2+0*i
 
  • #4
Benzoate said:
so Re (1/z) = Re(1/(x+iy))=Re(1/(x+iy)*(x-iy)/(x-iy))=Re(x-iy)/(x^2-y^2)). How will this new expression make it easier for me to mapped Re(1/z) in a complex plane. In addition, how would I mapped 1/2 on a complex plane? Does 1/2=1/2+0*i

First … it's Re(1/z) = Re(x-iy)/(x^2 + y^2), not x^2-y^2 :wink:

second … Re(x + iy) < 1/2 if … ? :smile:
 
  • #5
tiny-tim said:
First … it's Re(1/z) = Re(x-iy)/(x^2 + y^2), not x^2-y^2 :wink:

second … Re(x + iy) < 1/2 if … ? :smile:

Only if x less than or equal to 1/2 and y is less than or equal to zero

Wouldn't Re(x+iy/(x^2+y^2)) Be sketched only on the x-axis since all real numbers lie only on the x-axis. Would it be correct to say that 1/2 lies only only the real part of the complex plane since 1/2=1/2+i*0?
 
  • #6
Benzoate said:
Only if x less than or equal to 1/2 and y is less than or equal to zero

No … the value of y doesn't matter …

Re(x+iy) = x …

so Re(x+iy) < 1/2 just means x < 1/2 …

so Re(x+iy) < 1/2 for x < 1/2 and for all y. :smile:
Wouldn't Re(x+iy/(x^2+y^2)) Be sketched only on the x-axis since all real numbers lie only on the x-axis. Would it be correct to say that 1/2 lies only only the real part of the complex plane since 1/2=1/2+i*0?

Forget 1/2.

And you're not sketching Re(x+iy/(x^2+y^2)).

You're sketching x + iy such that Re(x+iy/(x^2+y^2)) ≤ 1/2.
 

1. What does it mean to "sketch the closure" of a set?

Sketching the closure of a set is the process of visualizing all the points that are contained within the set, as well as any limit points that may exist on the boundary of the set. This helps to understand the structure and boundaries of the set.

2. How is the closure of a set different from the set itself?

The closure of a set includes the set itself, along with any limit points that may exist on the boundary. This means that the closure is a larger set than the original set, as it includes additional points that are not necessarily in the set itself.

3. Why is it important to sketch the closure of a set?

Sketching the closure of a set allows us to understand the completeness and connections between different points in the set. It also helps to identify any limit points and visualize the boundaries of the set, which can be useful in various mathematical and scientific applications.

4. Can the closure of a set be empty?

Yes, in certain cases the closure of a set can be empty. This occurs when the set itself does not contain any limit points, and therefore there are no additional points to be included in the closure.

5. How can I determine the closure of a set?

To determine the closure of a set, you can start by identifying all the points that are contained within the set. Then, consider any limit points that may exist on the boundary of the set. Adding these limit points to the original set will give you the closure of the set.

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