Find complex number Z in $\lvert Z\rvert= Z+3-2i$ $$\lvert Z\rvert = Z+ 3-2i$$
what I did so far is
let $Z = a +bi$
so $$\sqrt{a^2 + b^2} = a+bi+3-2i$$
$$\sqrt{a^2 + b^2} = a+3 + i (b-2)$$
now what I'm thinking is squaring both sides but that doesn't work, any tips?
 A: Hint
Once $\sqrt{a^2+b^2}\in \Bbb R$ then $(a+3)+i(b-2)\in \Bbb R$ and then
$$\sqrt{a^2+b^2}=a+3\\
b-2=0$$
Can you finish?
A: Hint: from $z = |z| - 3 + 2i\,$, taking the complex conjugate of both sides gives $\bar z = |z| - 3 - 2i\,$. Then, multiplying the two and using that $z \bar z = |z|^2\,$:
$$
|z|^2 = \left(|z| - 3 + 2i\right)\left(|z| - 3 - 2i\right)
$$
The above is a linear equation in $|z|\,$. Once $|z|$ is determined, $z$ follows from the original relation.
A: Given that for $|z|$ is real, we know that $w:=z+3-2i$ is real, and so $\Im(w)=0$.
If $z=a+bi$, this means that $w=(a+3)+(b-2)i$. From before, we get that $b=2$.
From $|z|=w$ we have that $\sqrt{a^2+2^2}=a+3$.
A: Solve
$$ 
\lvert z \rvert = z + 3 - 2i
\tag{1}
$$

1 
Observation: $\lvert z \rvert$ is real, while the right hand side has the imaginary term $-2i$.

For the equality to hold, the imaginary part of $z$ must be $2i$ to cancel the imaginary component, that is
$$z=x+2i\tag{2}$$
2 
Use $(2)$ in $(1)$ to obtain
$$
 \lvert z \rvert = \lvert x + 2i \rvert = \sqrt{x^{2}+4} = x + 3
\tag{3}
$$
3
Solve $(3)$ to recover
$$
 x = -\frac{5}{6}
$$
4
The value of $z$ which solves $(1)$ is
$$
 z = -\frac{5}{6} + 2 i
$$

Confirmation
$$
\begin{align}
\require{cancel}
  \lvert z \rvert &= z + 3 - 2i \\[10pt]
  \Big\lvert -\frac{5}{6} + 2 i \Big\rvert &= \left(-\frac{5}{6} \cancel{+ 2 i} \right) + 3 \cancel{- 2i} \\[10pt]
  \sqrt{\frac{25}{36} + 4} &= -\frac{5}{6} + 3\\[10pt]
  \sqrt{\frac{169}{36}} &= \frac{13}{6}\\[10pt]
   \frac{13}{6} &= \frac{13}{6}
\end{align}
$$
