How do solve inequality $|\frac{z}{1+z}|<1$?

Question

How do I solve for $z$ in

$$\left|\frac{z}{1+z}\right|<1$$

I tried using $z= x + iy$ but that becomes complicated, any idea guys?

Answer 1

$$\left|\frac z{1+z}\right|=\frac{|z|}{|1+z|}<1$$

$$|z|<|1+z|$$

$$|z|^2<|1+z|^2$$

Let $z=x+iy$,

$$x^2+y^2<(1+x)^2+y^2$$

$$0<1+2x$$

$$x>-\frac12$$