Calculate and describe the whole complex numbers group which...

Question

Calculate and describe the set of complex numbers which:

A) $$\frac{1}{Z} + \frac{1}{Z} = 1$$

B) $$|\frac{Z - 1}{Z + 1}| <= 1$$

Which steps should I follow? Any advices?

Answer 1

1) Note that $z \neq 0$

${1 \over z} + {1 \over z} = {2 \over z} = 1$

Solving for z gives $z = 1/2$.

2)$|{{z-1} \over {z+1}}| \leq 1$

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

$(x-1)^2+y^2 \leq (x+1)^2+y^2$

$(x-1)^2\leq(x+1)^2$

$x^2 - 2x + 1 \leq x^2 + 2x + 1$

$-4x \leq 0$

$x \ge 0$

Answer 2

The equation (1) is $2/z=1$ which is satisfied by unique number that is 2. and provided $z \ne 0$

The second equation $$|z-1|<=|z-(-1)|$$

$$\implies |z-1|-|z-(-1)|<=0$$

The second question is asking about all points $(x,y)$ such that distance of point $(x,y)$ to $(1,0)$ minus the distance of point $(x,y)$ to $(-1,0)$ is less than 0.

If you draw points on the plane you will see that the equation is satisfied by all the points on right side of the y-axis. Hope this helped