# On the solution of the equation $z+ \alpha \left| z-1\right| + 2i = 0$

Question:-

Find the range of real number $\alpha$ for which the equation $z+ \alpha \left| z-1\right| + 2i = 0$; $z=x+iy$ has a solution. Also find the solution.

Attempt at a solution:-

On substituting $z=x+iy$ in the equation $z+ \alpha \left| z-1\right| + 2i = 0$, we get that the imaginary part of the complex variable $z$ should be $-2$, i.e $y=-2$. So, the complex variable is of the form $z=x-2i$.

As, the real part of the equation is also $0$, so we get the following equation for the real part

$$\begin{equation}x=-\alpha\sqrt{x^2-2x+5} \end{equation} \tag{1}$$

After squaring both sides of the equation, we arrive at the following quadratic equation.

$$\begin{equation} (\alpha^2-1)x^2-2\alpha^2x+5\alpha^2=0 \end{equation} \tag{2}$$

Now, as $x$ has to be real so $D \ge 0 \implies 5-4\alpha^2 \ge 0 \implies -\dfrac{\sqrt{5}}{2}\le \alpha \le \dfrac{\sqrt{5}}{2}$

Now, let's consider different cases.

Case 1:-

When $\alpha^2=1$, then from $(2)$, we get $x=\dfrac{5}{2}$. Now, after plugging $x=\dfrac{5}{2}$ in $(1)$, we get $\dfrac{5}{2}=-\alpha\left(\dfrac{5}{2}\right)$, so from this we conclude that when $x=\dfrac{5}{2}$, then, $\alpha=-1$

The place where I am getting stuck:-

I am not able to think up of the different solutions of $x$, when $\alpha^2 \neq 1$. If anyone can help me think of how to go about solving for the remaining cases.

• For a start, if $\alpha>0$ then we need $x<0$, if $\alpha<0$ we need $x>0$, and $\alpha = 0$ clearly works. – Joey Zou Jul 9 '16 at 4:21
• How did you get that – user350331 Jul 9 '16 at 4:35
• Look at equation (1). The square root (which came from the absolute value) is always nonnegative, and in this case it's always positive because we know $z-1$ can't be zero. – Joey Zou Jul 9 '16 at 4:38

## 1 Answer

From $(1)$, $\alpha=0$ works. If $\alpha\not=0$, then $$\sqrt{(x-1)^2+4}=\frac{x}{-\alpha}\gt 0\tag3$$

From $(2)$, for $-\frac{\sqrt 5}{2}\le \alpha\le\frac{\sqrt 5}{2}$ with $\alpha^2\not=1$, $$x=\frac{\alpha^2\pm \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}$$

Case 1 : For $\alpha=0$, $x=0$.

Case 2 : For $\alpha^2=1$, you've already done. ($\alpha=-1,x=5/2$)

Case 3 : For $-\frac{\sqrt 5}{2}\le \alpha\lt -1$, from $(3)$, $x\gt 0$, so $x=\frac{\alpha^2\color{red}{\pm} \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}$.

Case 4 : For $-1\lt \alpha\lt 0$, from $(3)$, $x\gt 0$, so $x=\frac{\alpha^2\color{red}{+} \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}$.

Case 5 : For $0\lt \alpha\lt 1$, from $(3)$, $x\lt 0$, so $x=\frac{\alpha^2\color{red}{+} \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}$.

Case 6 : For $1\lt \alpha\le\frac{\sqrt 5}{2}$, from $(3)$, $x\lt 0$, so there is no solution.

Therefore, the range of real number $\alpha$ for which the equation $z+ \alpha \left| z-1\right| + 2i = 0$; $z=x+iy$ has a solution is $\color{red}{-\frac{\sqrt 5}{2}\le\alpha\lt 1}$.

And the solution is $$\color{red}{x=\frac{\alpha^2\pm \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}\qquad\text{for \ -\frac{\sqrt 5}{2}\le\alpha\lt -1}}$$

$$\color{red}{x=\frac 52\qquad\text{for \ \alpha=-1}}$$ $$\color{red}{x=\frac{\alpha^2+ \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}\qquad\text{for \ -1\lt\alpha\lt 1}}$$

• So, all I had to do was see for what all values $\dfrac{x}{-\alpha}$ is positive. Am, I right. – user350331 Jul 9 '16 at 5:11
• @user350331: Yes, for each $\alpha\not=0$, you need to find $x$ such that $x/(-\alpha)$ is positive. – mathlove Jul 9 '16 at 5:12
• Would a geometrical approach be possible to this question. – user350331 Jul 9 '16 at 5:55
• @user350331: $|z-1|$ has a geometrical meaning, but on the right hand side of $\alpha|z-1|=-z-2i$ there is another $z$ (and treating $-z-2i$ geometrically seems difficult), which might make it impossible to have a geometrical approach if I'm not mistaken. – mathlove Jul 9 '16 at 6:07
• @user350331: Oh, sorry, I misunderstood your point. The denominator is $\alpha^2-1$, so I wanted to separate it into $\alpha^2-1\gt 0$ or $\alpha^2-1\lt 0$. For example, in case 5, $$x=(\alpha^2\color{red}{-}\alpha\sqrt{5-4\alpha^2})/(\alpha^2-1)\gt 0\iff \alpha^2-\alpha\sqrt{5-4\alpha^2}\lt 0\iff \alpha\lt \sqrt{5-4\alpha^2}\iff \alpha^2\lt 5-4\alpha^2\iff \alpha^2\lt 1$$so $x=(\alpha^2\color{red}{-}\alpha\sqrt{4-5\alpha^2})/(\alpha^2-1)$ is positive, and so this has to be eliminated. I did the other cases in the similar way as above. – mathlove Jul 9 '16 at 7:16