# Complex numbers: Find the Geometric image

1. Nov 30, 2014

### HMPARTICLE

1. The problem statement, all variables and given/known data

Find the Geometric image of;

1. $| z - 2 | - | z + 2| < 2;$
2. $0 < Re(iz) < 1$

2. Relevant equations

3. The attempt at a solution
In both cases i really am struggling to begin these questions, complex numbers are not my best field.

There are problems before this one like $| z - 1 + 2i | >3$ which is the exterior of a circle with center (1,-2) with radius 3.

I know that Re(z) is a function that gives the real component of complex number.

Just a push into the light please.

2. Nov 30, 2014

### PeroK

For 1) you might think about conic sections.

2) shouldn't be hard. Why not set z = x + iy? And see what comes out.

3. Dec 1, 2014

### HMPARTICLE

for the second one i get {(x,y) in R such that -1 < y < 0 }, according to the solutions, that is correct.

the first one, I am still stuck on.
When you say conic sections, im thinking hyperbola.
standard form of hyperbola is;

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
ignoring the advice, i tried to square both sides of the inequality.

i get

$(|z−2|−|z+2|)^2<4$
$| z-2|^2 - 2|z-2||z+2| + |z+2|^2 < 4$

do i continue in this fashion?

i get to this

$2(x^2 - y^2) + 8 - 2|z-2||z+2| < 4$

I could subtract 8 from both sides.
But then if i square both sides again i'm going to be left with something similar to |z-2||z+2| which i dont think i can do anything with.

I told you i'm TERRIBLE with the complex realm! haha

4. Dec 1, 2014

### Joffan

What is the solution of (1) on the real number line? Identify the point(s) where it is an equality not an inequality.

Then perhaps exploring the region of the complex plane nearby will give some clues.

Notice that -3 < 2 but 9 > 4, so your squaring may not have the results you want.

5. Dec 2, 2014

### HMPARTICLE

when z is -1 then the left hand side is equal to 2. still i am in the dark.

I'm sorry guys! really not seeing this one.

6. Dec 2, 2014