Determine the set of all complex number z satisfying following conditions

Question

I’m having some troubles of calculating complex numbers where I need to deal with absolute values and inequalities. Here is an example I’ve been working on but I get stuck

Re(2/z)+Im(4/z)<1

I use z=x+iy And find 1/z = (x-iy)/(x^2+y^2)

So

Re(2/z) = 2x/(x^2+y^2) and Im(4/z) = (-4y)/(x^2+y^2)

If I just forget about the inequality and make it equal to 1. I get:

2x-i4y = x^2+y^2

Which is a circumference with a complex radius. Now I’m not sure what to do or if I’m going in the correct direction.

I hope you can help me.

Answer 1

That factor of $i$ you added is spurious. You should get $2x-4y<x^2+y^2$, which is the inside of a circle of real radius.

Answer 2

Solve $$\frac{2x}{x^2+y^2}-\frac{4y}{x^2+y^2}<1$$

Answer 3

Your equation is slightly wrong: that $i$ should not be there. It should be $$2x - 4y = x^2 + y^2 $$ Now its radius is a real number, and you should have no trouble plotting it.

To solve the inequality, just check a point on the inside of the circle, and another point on the outside. That will tell you whether the inside is part of the solution, and whether the outside is part of the solution.