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


1 Answer 1


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 :

$$\color{red}{\text{For $\ 0\le a\le\frac 12,\ $ there is no solution}}$$ $$\color{red}{\text{For $\ a\gt \frac 12,\ z=\frac{4a+ \sqrt{4a^2+3}}{4(2a+1)(2a-1)}+\frac 14i$}}$$

  • $\begingroup$ I got a tighter bound than yours, though it produces the same answer, it is $x \ge \dfrac{3}{8a}$, which we get it from $4ax=1+2\sqrt{x^2+16}$, as least value of $x^2$ is $0$. And thanks for another solution. $\endgroup$
    – user350331
    Jul 9, 2016 at 15:07
  • $\begingroup$ @user350331: You are welcome. (by the way, how did you get the equation in your comment? Just a typo? It is not the same as $(2)$ in my answer.) $\endgroup$
    – mathlove
    Jul 9, 2016 at 15:11
  • $\begingroup$ Oops, it's a typo it is meant to be $4ax=1+2\sqrt{x^2+\dfrac{1}{16}}$ $\endgroup$
    – user350331
    Jul 9, 2016 at 15:15

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