Solve the complex number equation $|z|-\bar{z}=i$ Solve the complex number equation $|z|-\bar{z}=i$.
The following was my thought process:
$$|z|=i+\bar{z}$$
Given that $Im(|z|)=0$, $Im(\bar{z})=-i$. Hence, $\bar{z}=a-i$ and $z=a+i$.
$$|z|=i+(a-i)$$
$$|z|=a$$
But since $z=a+i$, $|z|=\sqrt{a^2+1}$ and hence I've reached a contradiction, as $\sqrt{a^2+1}\neq a$.
Hence, there is no solution for $|z|-\bar{z}=i$, for $z\in\mathbb{C}$.
 A: You havent gone wrong anywhere. The contradiction tells that there does not exist any $z\in \mathbb{C}$  satisfying your condition.
A: The posted solution is correct. Another way is to write it as $\,\bar z = |z| - i\,$, then take the conjugates on both sides $\,z = |z| + i\,$ and multiply the two to get $\,|z|^2=|z|^2 + 1\,$, so no solutions exist.
A: Your approach is a good one and your reasoning is correct.
A: $$\sqrt{a^2+b^2}-a+ib=i$$ implies $b=1$, so $\sqrt{a^2+1}=a$ by identification.
A: Nothing wrong with yours. Had an idea for an alternative.
$|z|=i+\bar{z}$
$|z|=-i+z$ by cojugating both sides.
$|z|=(z+\bar{z})/2$ by addition and halfing both sides.
The modulus is therefore equal to the real part, so the imaginary part must be zero. This means z is itself real.
Let $z=x$, $x$ is real.
Returning to the first expression, this implies
$x=i+x\implies i=0$, a contradiction.
A: 
Note: I used $z^\star$ because I could not do the conjugate bar on my drawing
Here is a visual proof:
$|z|$ and $\bar z$ belong to the same black circle because $|z|=|\bar z|=\Big\lvert|z|\Big\lvert$, therefore $|z|-\bar z$ belongs to the blue circle which never crosses the $Y-$axis except in $0$.
In particular it cannot pass through $i$.
A: Well, let's solve a more general problem. Suppose $\text{z}_1\space\text{z}_2\in\mathbb{C}$, solve:
$$\left|\text{z}_1\right|-\overline{\text{z}_1}=\text{z}_2\tag1$$
Let's assume $\text{z}_1=\Re\left(\text{z}_1\right)+\Im\left(\text{z}_1\right)i$ and $\text{z}_2=\Re\left(\text{z}_2\right)+\Im\left(\text{z}_2\right)i$, so we get:
$$\left|\Re\left(\text{z}_1\right)+\Im\left(\text{z}_1\right)i\right|-\overline{\Re\left(\text{z}_1\right)+\Im\left(\text{z}_1\right)i}=\Re\left(\text{z}_2\right)+\Im\left(\text{z}_2\right)i\tag2$$
Rewriting gives:
$$\sqrt{\Re\left(\text{z}_1\right)^2+\Im\left(\text{z}_1\right)^2}-\left(\Re\left(\text{z}_1\right)-\Im\left(\text{z}_1\right)i\right)=\Re\left(\text{z}_2\right)+\Im\left(\text{z}_2\right)i\tag3$$
So, we can write a system of equations:
$$
\begin{cases}
\sqrt{\Re\left(\text{z}_1\right)^2+\Im\left(\text{z}_1\right)^2}-\Re\left(\text{z}_1\right)=\Re\left(\text{z}_2\right)\\
\\
\Im\left(\text{z}_1\right)=\Im\left(\text{z}_2\right)
\end{cases}\tag4
$$
Solving for $\Re\left(\text{z}_1\right)$, gives:
$$\Re\left(\text{z}_1\right)=\frac{\Im\left(\text{z}_1\right)^2-\Re\left(\text{z}_2\right)^2}{2\Re\left(\text{z}_2\right)}\tag5$$

Now, for your case it is not hard to see that we have a problem when $\Re\left(\text{z}_2\right)\to0$ for $(5)$.

