Finding $\{z\in \mathbb{C}\mid|1 + i z|^2<1,\,\mathrm{Im}(z-i)>0\}$

Question

I am looking for the following set:

$$\{z\in \mathbb{C}\mid|1 + i z|^2<1,\,\mathrm{Im}(z-i)>0\}$$

What I have done so far:

$z= x+iy, \quad x,y \in \mathbb{R} , i \in \mathbb{R}$

$ \begin{align} |1+iz|^2<1 & \equiv |1+ i(x+iy)|^2 <1 \\ &\equiv |1+ix+i^2y|^2 < 1\\ &\equiv |1+ix-iy|^2 < 1\\ \end{align} $

How could I proceed to find the solution set? Am I on the right track so far?

Answer 1

with $z=a+ib$ we get$$|1-b+ai|^2=(\sqrt{(1-b)^2+a^2})^2<1$$ and $$\Im(z-i)=b-1>0$$

Answer 2

Observe that $$|iz+1|=|i|\left|z+\dfrac{1}i\right|=|z-i|$$ and $|z-i|^2\lt1$ is same as $$\color{Green}{(|z-i|+1)(|z-i|-1)}=|z-i|^2-1^2\lt 0.$$ Moreover among those colored factors first one is positive.
Therefore we need the circle $|z-i|\lt 1$ intersect with $\Im(z-i)\gt 0, $ which is the open upper half disk with radius one centered at $i.$

Answer 3

Note that $$|1+iz|^2=|1+i(x+iy)|^2=|1+ix-y|^2 =|(1-y)+ix|^2=(1-y)^2+x^2.$$ Then, $$|1+iz|^2<1 \Leftrightarrow (1-y)^2+x^2<1,$$ i.e., the set is the interior of a unitary disc centered in $P(0,1)$.

And the set where $Im (z-i) =Im(x+iy-i)=Im(x+i(y-1))=y-1>0 \Leftrightarrow y>1$.

Then, ploting the two inequalities in the same cartesian system os coordinates, the intercet of two conditions is the interior of a semi-disc construct above.