To find the solution of the equation $2\left|z \right|-4az+1+ia=0$ Question:-
For every real number $a \ge 0$, find all the complex numbers $z$, satisfying the equation $2\left|z \right|-4az+1+ia=0$

Attempt at a solution:-
Let $z=x+iy$, then the equation $2\left|z \right|-4az+1+ia=0$ becomes as follows
$$\begin{equation}
2\sqrt{x^2+y^2}-4a(x+iy)+1+ia=0 \\ 
\left(2\sqrt{x^2+y^2}-4ax+1\right)+i(a-4ay)=0+i\cdot0
\end{equation}$$
Now, equating the imaginary part and real part on both sides, we get
$$\begin{equation}
y=\dfrac{1}{4} \qquad \left(a\neq0\right)
\end{equation}$$
$$\begin{equation}
2\sqrt{x^2+y^2}-4ax+1=0 \tag{1}
\end{equation}$$
Putting $y=\dfrac{1}{4}$ in $(1)$, we get 
$$2\sqrt{x^2+\dfrac{1}{16}}-4ax+1=0 \implies 2\sqrt{x^2+\dfrac{1}{16}}=4ax-1$$
On squaring both sides we get the roots of $x$ as $$x=\dfrac{4a\pm\sqrt{4a^2+3}}{4(4a^2-1)}$$

The place where I got stuck:-
From here onward I am not able to think anything
 A: For $a=0$, there is no solution. In the following, $a\gt 0$.
You have
$$2\sqrt{x^2+\dfrac{1}{16}}=4ax-1\tag2$$
Here, note that you need to have
$$4ax-1\gt 0\iff x\gt \frac{1}{4a}\tag3$$
Squaring the both sides of $(2)$ gives
$$(64a^2-16)x^2-32ax+3=0\tag4$$
Case 1 : For $64a^2-16=0$, i.e. $a=1/2$, $x=3/16$ but this does not satisfy $(3)$.
Case 2 : For $64a^2-16\not=0$, 
$$x=\frac{4a\pm \sqrt{4a^2+3}}{16a^2-4}$$
Now we consider the following inequality :
$$x=\frac{4a\pm \sqrt{4a^2+3}}{4(2a+1)(2a-1)}\gt \frac{1}{4a}\tag5$$
Case 2-1 : For $2a-1\gt 0$, i.e. $a\gt 1/2$,
$$(5)\iff 4a^2\pm a\sqrt{4a^2+3}\gt 4a^2-1\iff \pm a\sqrt{4a^2+3}\gt -1$$
Now $a\sqrt{4a^2+3}\gt -1$ holds.
$$- a\sqrt{4a^2+3}\gt -1\iff a\sqrt{4a^2+3}\lt 1$$
This does not hold since $a\gt 1/2$.
So, in this case, 
$$x=\frac{4a\color{red}{+} \sqrt{4a^2+3}}{4(2a+1)(2a-1)}$$
Case 2-2 : For $2a-1\lt 0$, i.e. $a\lt 1/2$. Similarly, we get
$$(5)\iff 4a^2\pm a\sqrt{4a^2+3}\lt 4a^2-1\iff \pm a\sqrt{4a^2+3}\lt -1$$
Now $a\sqrt{4a^2+3}\lt -1$ does not hold.
$$-a\sqrt{4a^2+3}\lt -1\iff a\sqrt{4a^2+3}\gt 1$$
This does not hold since $a\lt 1/2$.
So, in this case, there is no solution.
Therefore, the answer is the following :
$$\color{red}{\text{For $\ 0\le a\le\frac 12,\ $ there is no solution}}$$
$$\color{red}{\text{For $\ a\gt \frac 12,\  z=\frac{4a+ \sqrt{4a^2+3}}{4(2a+1)(2a-1)}+\frac 14i$}}$$
