Solve the complex number equation $|z|+z=1$ Solve the equation $|z|+z=1$.
This is my thought process so far:
$$|z|=1-z$$
$$\sqrt{a^2+b^2}=1-z$$
$$a^2+b^2=1-2z+z^2$$
$$a^2+b^2=1-2(a+bi)+(a+bi)^2$$
$$a^2+b^2=1-2a-2bi+a^2+2abi-b^2$$
$$a^2+2b^2=1-2a-2bi+a^2+2abi$$
$$2b^2=1-2a-2bi+2abi$$
Not sure what quite to do after this stage. Perhaps I am approaching this question wrongly.
Edit: the book I'm picking this question up from clearly states "Solve for $z\in\mathbb{c}$", so perhaps is there another solution with an imaginary part as well?
 A: $$\sqrt{a^2+b^2}+a+ib=1$$ immediately implies $b=0$ by identification of the imaginary part(s).
You preferred a harder way, with squaring,
$$a^2+b^2=1-2a-2ib+a^2-b^2+2iab$$ and again by identification
$$-2b+2ab=0$$
yields $b=0$ or $a=1$.
Now plugging $a=1$,
$$1+b^2=-b^2$$ is a dead end.
A: Another way is to write it as $\,z = 1 - |z|\,$, then compare the magnitudes on the two sides where the RHS is a real number, so  $\,\require{cancel} \bcancel{|z|^2} = \left(1 - |z|\right)^2 = 1 - 2 |z| + \bcancel{|z|^2} \iff |z| = \frac{1}{2}\,$.
Substituting back into the original equation gives $\,z = 1 - |z| = 1 - \frac{1}{2} = \frac{1}{2}\,$.

[ EDIT ] $\;$ More generally, the same approach can be used to solve $|z|+z=\alpha$ for arbitrary $\alpha \in \mathbb C$.

*

*Isolate $z$ to one side: $\;z = \alpha - |z|\,$.


*Conjugate: $\;\bar z = \overline\alpha - |z|\,$.


*Multiply the two: $\;\bcancel{|z|^2} = \big(\alpha - |z|\big)\big(\overline\alpha-|z|\big)=|\alpha|^2 + \bcancel{|z|^2} - 2 |z| \text{Re}(\alpha)\,$. If $\,\text{Re}(\alpha) = 0\,$ no solutions exist unless $\,\alpha = 0\,$, and in that case the solution set is $\,z \in \mathbb R^{\le 0}\,$.


*Otherwise, solve for $|z|$: $\;|z| = \frac{|\alpha|^2}{2 \text{Re}(\alpha)}\,$. If $\,\text{Re}(\alpha) \lt 0\,$ no solutions exist, since it would imply $\,|z| \lt 0\,$.


*Otherwise, substitute back into the original equation: $\;z = \alpha-|z|=\alpha-\frac{|\alpha|^2}{2 \text{Re}(\alpha)}\,$.
OP's problem is the case $\,\alpha=1\,$, so $\,z = 1 - \frac{|1|^2}{2 \cdot 1}=\frac{1}{2}\,$.
