# Homework Help: Sketching the closure of a set

1. Aug 31, 2008

### Benzoate

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

Sketch the closure of the set:Re(1/z)=< 1/2

[ b]2. Relevant equations[/b]

3. 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?

2. Aug 31, 2008

### tiny-tim

Hi Benzoate!

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).

3. Aug 31, 2008

### Benzoate

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. Aug 31, 2008

### tiny-tim

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

second … Re(x + iy) < 1/2 if … ?

5. Aug 31, 2008

### Benzoate

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. Aug 31, 2008

### tiny-tim

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.
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.