Find set z where $f(z)$ expand the area

Question

$f(z) = 2z + iz^2 \Rightarrow f'(z) = 2 + 2iz$
So I need to find where $|f'(z)|>1$. And I really confused with it
My attempt: let's use the formula $z\bar z = |z|^2 \Rightarrow 4(1+iz)(1-iz)=4(1+z^2)>1 \Rightarrow z^2 > -\frac{3}{4}$. Here I'm stuck, 'cause next step $|z| > i\frac{\sqrt{3}}{2}$ doesn't make sense, as $|z|$ supposed to be from $\mathbb{R}$. And I can't understand the step where I do incorrect things

Answer 1

$1<|f'(x)|=|2+2iz|=|2i|\cdot|-i+z|=2|z-i|$.

Therefore, the inequality is equivalent to $|z-i|>1/2$. These are the points at distance $>1/2$ from the point $i$: the outside of the circle with center $i$ and radius $1/2$.