Complex Number Inequalities: Sketching Solutions

[nayfie]

Hey guys,

Just having a bit of trouble with inequalities.

Homework Statement

Sketch all complex numbers 'z' which satisfy the given condition:

$$|z + i + 1| \leq |z - i|$$

Homework Equations

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The Attempt at a Solution

$$z + i + 1\leq z - i$$

$$z + 2i + 1\leq z$$

$$2i + 1\leq 0$$

$$i\leq\frac{1}{2}$$

No matter how hard I try, I can't seem to be able to solve for z.

Any help would be appreciated.

Thank you!

 

[gneill]

z is a complex number. So write it as z = a + b*i, where i is the square root of negative one. Put that into your inequality and see what happens. There will be some necessary relationship between a and b that satisfies the inequality.

 

[Lavabug]

Tried it myself, you sub z for x + iy as gneill said, regroup terms again in x + iy form and calculate the modulus on both sides, terms will cancel. I got the relationship: y <= -1/2 -x.

Looks like all numbers underneath and including the line y = -x -1/2 if I'm not mistaken?

Edit: One of the recommended threads: https://www.physicsforums.com/showthread.php?t=298041 Same problem basically.

 

[nayfie]

Firstly, thanks for your contribution gneill and Lavabug.

I have put together what you two have written and have come to an answer, which I'll add here in case somebody has a similar problem.

$$|z + i + 1|\leq |z - i|$$
$$|x + yi + i + 1|\leq |x + yi - i|$$
$$|x - x + yi - yi + i + 1|\leq |-i|$$
$$x + x + yi + yi + i + 1\leq i$$
$$2x + 2yi + 1\leq 0$$
$$2x + 2yi\leq -1$$
$$x + yi\leq\frac{-1}{2}$$
$$z\leq\frac{-1}{2}$$

As far as I can tell that solution is correct. Feel free to correct me if it's not.

 

[gneill]

For an imaginary number x + yi, the modulus (or magnitude) is
$$|x + y i| = \sqrt{x^2 + y^2}$$
Note also that |i| = |-i| = 1

Note that you can't just bring variables or constants out the modulus on one side of the equation and put them into the modulus on the other. So your third line doesn't make sense (and if you performed the obvious operations inside the modulus you'd see that the x's and yi's would cancel, leaving |i + 1| <= |-i|, which is clearly false.

You should replace z with x + yi as you've done, but expand the modulus on either side. Square root of the sum of the squares of the real and imaginary parts for each.

 

[I like Serena]

Hi nayfie!


I see you were only asked to sketch the result of the inequality.

Perhaps it would be useful to realize that complex numbers behave mostly like points in the X-Y-plane.
The absolute value of a complex number is defined as the distance to the origin in the X-Y-plane.

The solution to your inequality are those points in the X-Y-plane that are closer to 1 specific point than another specific point.

That reduces your problem to determining which points those are.


Cheers!