Complex number inequality graph

AI Thread Summary
The discussion focuses on how to graph the inequality Re(z) < 0 in the complex plane, specifically noting that it should represent the left half-plane where the real part of z is negative. Participants express confusion about the relationship between different sets and how to visualize the graph correctly. The conversation highlights the importance of understanding the definitions of the sets involved, particularly alpha and beta, and their independence from each other. Clarifications are made regarding the intersection of these sets and how to approach graphing them separately. Ultimately, the participants arrive at a clearer understanding of the graphical representation of Re(z) < 0.
Cpt Qwark
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Homework Statement


How would Re(z)<0 be graphed?

Homework Equations


Re(z) is the real part of z

The Attempt at a Solution


It looks similar to y>x, but only shaded in the third quadrant, how can this be explained? not relevant anymore
 
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Is there more to the problem?

Let z=x+iy, where x and y are real.
What is Re(z)?
 
The question doesn't tell, but let's just assume z=x+iy.
 
Cpt Qwark said:

Homework Statement


How would Re(z)<0 be graphed?

Homework Equations


Re(z) is the real part of z

The Attempt at a Solution


It looks similar to y>x, but only shaded in the third quadrant, how can this be explained?
I'm not sure I understand. It should be half the plane.
 
I think i might be getting confused with the set notation,

if \alpha=\{z||x+iy|&lt;n\} and \beta=\{z|Re(z)&gt;0\}, to graph \beta would it just be the real part of \alpha?
\{x,y,n\} \in \mathbb{R}
 
Cpt Qwark said:
I think i might be getting confused with the set notation,

if \alpha=\{z||x+iy|&lt;n\} and \beta=\{z|Re(z)&gt;0\}, to graph \beta would it just be the real part of \alpha?
\{x,y\} \in \mathbb{R}
Your definitions of alpha and beta are quite independent of each other, so why should there be any relationship between them?
Please post the question exactly as given.
 
take \alpha=\{z||e^{iπ/4}(z+2)|&lt;2\} (|e^{iπ}|=1) and \beta=\{z|Re(z)&gt;0\},
how would beta be graphed?
 
Cpt Qwark said:
take \alpha=\{z||e^{iπ/4}(z+2)|&lt;2\} (|e^{iπ}|=1) and \beta=\{z|Re(z)&gt;0\},
how would beta be graphed?
That cannot be the whole question since the information about alpha is irrelevant.
 
haruspex said:
That cannot be the whole question since the information about alpha is irrelevant.
sorry, forgot graph\alpha∩\beta
 
  • #10
Cpt Qwark said:
sorry, forgot graph\alpha∩\beta
That makes a huge difference!
Ok, so what do you get for alpha and beta separately?
How would you combine the graphs to get the intersection?
 
  • #11
haruspex said:
That makes a huge difference!
Ok, so what do you get for alpha and beta separately?
How would you combine the graphs to get the intersection?

I can graph alpha which is just |z+2]&lt;2 but for beta is it just the real part of alpha?
 
  • #12
Cpt Qwark said:
I can graph alpha which is just |z+2]&lt;2 but for beta is it just the real part of alpha?
No, the graph of beta has no connection with alpha. Why do you think it should? Forget alpha for the moment and figure out what beta looks like.
Or, did you mean to ask whether the intersection of the two is just the real part of alpha?
 
  • #13
haruspex said:
No, the graph of beta has no connection with alpha. Why do you think it should? Forget alpha for the moment and figure out what beta looks like.
Or, did you mean to ask whether the intersection of the two is just the real part of alpha?

My original question was that I'm not sure how Re(z)>0 is graphed when there is no relation to alpha.
 
  • #14
Cpt Qwark said:
My original question was that I'm not sure how Re(z)>0 is graphed when there is no relation to alpha.
Ok. In the z=x+iy formulation, what does Re(z) look like in terms of x and y? What region of the complex plane does that correspond to?
 
  • #15
haruspex said:
Ok. In the z=x+iy formulation, what does Re(z) look like in terms of x and y? What region of the complex plane does that correspond to?
Oh, ok I get it now.
Thanks for your help!
 

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