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Solving Inequality With Complex Numbers Question

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"Solving Inequality With Complex Numbers" Question

Homework Statement



What does the inequality pz + conjugate(pz) + c < 0 represent if |p|^2 >c ?

Homework Equations



p is a constant and a member of the set of complex numbers. c is a constant and a member of the set of real numbers.

The Attempt at a Solution



First, for the EQUALITY pz + conjugate(pz) + c = 0, where p = a+ib, z = x+iy :

pz + conjugate(pz) + c = 0 => (a+ib)(x+iy) + conjugate((a+ib)(x+iy)) + c = 0 => ax + iay + ibx - by + ax - iay -ixb -by + c = 0 => y=((a/b)*x)+ c/(2*b).

So, the inequality is y> ((a/b)*x)+ c/(2*b).

MY QUESTION IS: WHAT IS THE IMPORTANCE OF --> |p|^2 >c WHEN SHOWING THE INEQUALITY --> pz + conjugate(pz) + c < 0.

I know that |p| = sqrt((a^2) + (b^2)), so |p|^2 >c implies ((a^2) + (b^2)) > c but I don't understand how this helps to show the the inequality pz + conjugate(pz) + c < 0... or more so, the inequality y > ((a/b)*x)+ c/(2*b).

ALSO, if I were to graph this inequality, would it be a (diagonal) dotted line (I say dotted because points on the line are not a part of the solution) where the graph is shaded above the dotted line?

Any help would be greatly appreciated.
Thank you.
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement



What does the inequality pz + conjugate(pz) + c < 0 represent if |p|^2 >c ?
It doesn't represent anything because the complex numbers do NOT form an "ordered field". There is no way to define "<" for complex numbers consistent with their arithmetic.

Homework Equations



p is a constant and a member of the set of complex numbers. c is a constant and a member of the set of real numbers.

The Attempt at a Solution



First, for the EQUALITY pz + conjugate(pz) + c = 0, where p = a+ib, z = x+iy :

pz + conjugate(pz) + c = 0 => (a+ib)(x+iy) + conjugate((a+ib)(x+iy)) + c = 0 => ax + iay + ibx - by + ax - iay -ixb -by + c = 0 => y=((a/b)*x)+ c/(2*b).

So, the inequality is y> ((a/b)*x)+ c/(2*b).

MY QUESTION IS: WHAT IS THE IMPORTANCE OF --> |p|^2 >c WHEN SHOWING THE INEQUALITY --> pz + conjugate(pz) + c < 0.

I know that |p| = sqrt((a^2) + (b^2)), so |p|^2 >c implies ((a^2) + (b^2)) > c but I don't understand how this helps to show the the inequality pz + conjugate(pz) + c < 0... or more so, the inequality y > ((a/b)*x)+ c/(2*b).

ALSO, if I were to graph this inequality, would it be a (diagonal) dotted line (I say dotted because points on the line are not a part of the solution) where the graph is shaded above the dotted line?

Any help would be greatly appreciated.
Thank you.
 
  • #3
I like Serena
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It doesn't represent anything because the complex numbers do NOT form an "ordered field". There is no way to define "<" for complex numbers consistent with their arithmetic.
That doesn't seem to be quite right, since the resulting expression is a real number.
And for real numbers the "<" inequality is defined.

pz + conjugate(pz) = 2 Re(pz)
 
  • #4
I like Serena
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I don't see any particular relevance either to |p|2 > c.

So, the inequality is y> ((a/b)*x)+ c/(2*b).
Not quite.
The direction of the inequality depends on the sign of b.
If b is negative, the inequality is inverted.


And I agree that you get a shaded half surface bounded by your dotted line.
You only have to be careful, which half surface you have exactly.
 
  • #5
HallsofIvy
Science Advisor
Homework Helper
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That doesn't seem to be quite right, since the resulting expression is a real number.
And for real numbers the "<" inequality is defined.

pz + conjugate(pz) = 2 Re(pz)
Right, I completely missed that.
 

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