Fing the range of alpha for equation to have solution?

**fardeen_gen** · May 23rd 2009, 09:30 PM

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

**Opalg** · May 24th 2009, 11:55 AM

Originally Posted by fardeen_gen

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

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

**fardeen_gen** · May 25th 2009, 09:48 AM

Originally Posted by Opalg

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

Am I doing it correct if I say that, for the equation to have a solution, $-\frac{\sqrt{5}}{2}\leq \alpha \leq \frac{\sqrt{5}}{2}$ and the solution in that case is:
$Z = \frac{2\alpha \pm \alpha\sqrt{5 - 4\alpha^2}}{\alpha^2 - 1} - 2i,\ \alpha\neq \pm 1;\ Z = \frac{5}{2} - 2i, \alpha = \pm 1$

**Opalg** · May 25th 2009, 11:23 AM

Originally Posted by fardeen_gen

Am I doing it correct if I say that, for the equation to have a solution, $-\frac{\sqrt{5}}{2}\leq \alpha \leq \frac{\sqrt{5}}{2}$ and the solution in that case is:
$Z = \frac{2\alpha \pm \alpha\sqrt{5 - 4\alpha^2}}{\alpha^2 - 1} - 2i,\ \alpha\neq \pm 1;\ Z = \frac{5}{2} - 2i, \alpha = \pm 1$

Yes, the condition for a solution to exist is $-\frac{\sqrt{5}}{2}\leq \alpha \leq \frac{\sqrt{5}}{2}$ (I carelessly wrote $2/\sqrt5$ in my previous comment—now corrected—where it should have been $\sqrt5/2$ .) But you should check your answer to the quadratic equation again. I get $z = \frac{\alpha^2 \pm \alpha\sqrt{5-4\alpha^2}}{\alpha^2-1}$ .

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