Find complex $z$ that satisfies two complex inequations

Question

I don't know if I am doing this right, can you help me with a hint?

$$ {|z-1| \over |z| } < {1} ; $$ and $$ {|z| \over |z-2| } < {1}$$ Find all the complex number $$z = x+iy$$ that satisfies these inequalities.

Solving for the first one: $$ {|z-1| \over |z| } < {|z-1| } < {|z|} < (z+1)^2 < z^2 $$ then $$ z^2+2z+1 <z^2 $$ Solving further I got that $$ z<-1/2 $$

For the second inequality I made a similar approach and I got that $ -1/2 < z < 1/2$

Can you help me with this? Thanks! :)

Answer 1

$${|z-1| \over |z| } < {1} \Leftrightarrow \\ |z-1| < |z| \Leftrightarrow \\ |z-1|^2 < |z|^2 \Leftrightarrow \\ (z-1)(\overline{z-1}) < z \overline{z} \Leftrightarrow \\ z \bar{z} -z -\bar{z}+1 < z \overline{z} \Leftrightarrow \\ 1 < z+ \bar{z} $$

Now, if $z= x+iy$ then $z+\bar{z}=2x$.

The second inequality can be solved the same way.

Answer 2

You are wrong starting from $$|z-1| < |z|$$ Instead it should be $$(x-1)^2+y^2 < x^2+y^2$$ $$(x-1)^2+y^2 < x^2+y^2$$ $$x > \frac{1}{2}$$

Answer 3

$$\frac{\mid z-1\mid}{\mid z\mid}<1\Leftrightarrow \frac{\sqrt{(x-1)^2+y^2}}{\sqrt{x^2+y^2}}<1\Leftrightarrow \sqrt{1+\frac{1-2x}{x^2+y^2}}<1\Leftrightarrow 1+\frac{1-2x}{x^2+y^2}<1\Leftrightarrow\frac{1-2x}{x^2+y^2}<0\Leftrightarrow 1-2x<0\Leftrightarrow 1<2x\Leftrightarrow \frac{1}{2}<x$$

$$\frac{\mid z\mid}{\mid z-2\mid}<1\Leftrightarrow\frac{\sqrt{x^2+y^2}}{\sqrt{(x-2)^2+y^2}}<1\Leftrightarrow \sqrt{x^2+y^2}<\sqrt{(x-2)^2+y^2}\Leftrightarrow x^2+y^2<x^2-4x+4+y^2\Leftrightarrow 4x<4\Leftrightarrow x<1$$

Putting these together yields $\frac{1}{2}<x<1$

Concerning your approach: $\frac{\mid z-1\mid}{\mid z\mid}<\mid z-1\mid$ only holds iff $\mid z\mid>1$, which is not necessarily the case.

$\mid z-1\mid<\mid z\mid$ indeed holds, cause it is equivalent to the condition $\frac{\mid z-1\mid}{\mid z\mid}<1$

$\mid z\mid<(z+1)^2$ however is wrong. $(z+1)^2$ is the square of a complex number and thus it can also be a complex number itself. The complex numbers aren't totally ordered, so you can't put them in inequalities like you can do with real quantities like $\mid z\mid$ and that $(z+1)^2<z^2$ doesn't hold follows subsequently (this wouldn't even be true if $z$ were real).

Answer 4

Hint:

For$z\ne 0$ the inequality becomes $$ |z-1|<|z| $$ this means that $z$ are represented by the points, in the Argand's plane, such that the distance from $A=(1,0)$ is less than the distance from $O=(0,0)$, that is the points on the right of the orthogonal bisector line of the segment $OA$, that is the line $x=\frac{1}{2}$.

You can use the same reasoning to solve the second inequality that is $$ |z|<|z-2| $$ and find the points on the left of the line $x=1$.

Now take the intersection of these two sets.

The figure is a further help ( blue $|z-1|<|z|$, red $|z|<|z-2|$)

Answer 5

According to me the best way in such questions is to first understand geometrically what is bring asked. Here you have the following iequalities

1.) $|z-1| < |z-0| $ i.e points in plane closer to to 1 than 0, hence to the right of the line $x=1/2$

2.) $|z-1| < |z-2|$ interpret it like above. Satisfied by points to left of $x=3/2$

3.) $|z-0| < |z-2|$ Satisfied by points to the left of $x=1$.

Points simultaneously satisfying all there are between $x=1/2$ and $x=1$

This is one of those problems where writing the solution took more time than getting it. I just drew the plane infront of me; marked the lines $x=0,1,2$, and marked the region above to get the solution.

Answer 6

The two inequalities can be viewer as the intersection of the interiors of the two hyperbolaes when we remove the points $A(0,0)$ and $B'(0,2)$

• $\mathbb E_1$ with centers $A(0,0), \, B(0,1)$

• and $\mathbb E_2$ with centers $A'(0,0), \, B'(0,2)$.

Obviously we say (draw a picture) that $\mathbb E_1 \subset \mathbb E_2$ which yields $\mathbb E_1\setminus {A(0,0)} \cap \mathbb E_2 \setminus {B'(0,2)}=\mathbb E_2\setminus {A(0,0)} $.

The remainder is trivial.