# Solving Inequality With Complex Numbers Question

• sarahs52

#### sarahs52

"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.

## 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.

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)

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.

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.