# Thread: Fing the range of alpha for equation to have solution?

1. ## Fing the range of alpha for equation to have solution?

Find the range of real number $\displaystyle \alpha$ for which the equation $\displaystyle z + \alpha|z - 1| + 2i = 0$ ; $\displaystyle z = x + iy$, has no solution. Also find the solution.

2. Originally Posted by fardeen_gen
Find the range of real number $\displaystyle \alpha$ for which the equation $\displaystyle z + \alpha|z - 1| + 2i = 0$ ; $\displaystyle z = x + iy$, has no solution. Also find the solution.
If $\displaystyle \alpha$ is real then $\displaystyle z+2i = -\alpha|z-1|$, which is real. So the imaginary part of z must be –2i. Write z = x–2i. Then $\displaystyle x = -\alpha|(x-1)-2i|$ and so $\displaystyle x^2 = \alpha^2\bigl((x-1)^2+4\bigr)$. The condition for that quadratic to have real roots is $\displaystyle \alpha^2\leqslant 5/4$. So the original equation has no solution for z if $\displaystyle |\alpha|>\sqrt5/2$.

3. Originally Posted by Opalg
If $\displaystyle \alpha$ is real then $\displaystyle z+2i = -\alpha|z-1|$, which is real. So the imaginary part of z must be –2i. Write z = x–2i. Then $\displaystyle x = -\alpha|(x-1)-2i|$ and so $\displaystyle x^2 = \alpha^2\bigl((x-1)^2+4\bigr)$. The condition for that quadratic to have real roots is $\displaystyle \alpha^2\leqslant 4/5$. So the original equation has no solution for z if $\displaystyle |\alpha|>2/\sqrt5$.
Am I doing it correct if I say that, for the equation to have a solution, $\displaystyle -\frac{\sqrt{5}}{2}\leq \alpha \leq \frac{\sqrt{5}}{2}$ and the solution in that case is:
$\displaystyle Z = \frac{2\alpha \pm \alpha\sqrt{5 - 4\alpha^2}}{\alpha^2 - 1} - 2i,\ \alpha\neq \pm 1;\ Z = \frac{5}{2} - 2i, \alpha = \pm 1$

4. Originally Posted by fardeen_gen
Am I doing it correct if I say that, for the equation to have a solution, $\displaystyle -\frac{\sqrt{5}}{2}\leq \alpha \leq \frac{\sqrt{5}}{2}$ and the solution in that case is:
$\displaystyle Z = \frac{2\alpha \pm \alpha\sqrt{5 - 4\alpha^2}}{\alpha^2 - 1} - 2i,\ \alpha\neq \pm 1;\ Z = \frac{5}{2} - 2i, \alpha = \pm 1$
Yes, the condition for a solution to exist is $\displaystyle -\frac{\sqrt{5}}{2}\leq \alpha \leq \frac{\sqrt{5}}{2}$ (I carelessly wrote $\displaystyle 2/\sqrt5$ in my previous comment—now corrected—where it should have been $\displaystyle \sqrt5/2$.) But you should check your answer to the quadratic equation again. I get $\displaystyle z = \frac{\alpha^2 \pm \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}$.