Find the range of real number for which the equation ; , has no solution. Also find the solution.
If is real then , which is real. So the imaginary part of z must be –2i. Write z = x–2i. Then and so . The condition for that quadratic to have real roots is . So the original equation has no solution for z if .
Last edited by Opalg; May 25th 2009 at 11:16 AM.
Reason: see below
If is real then , which is real. So the imaginary part of z must be –2i. Write z = x–2i. Then and so . The condition for that quadratic to have real roots is . So the original equation has no solution for z if .
Am I doing it correct if I say that, for the equation to have a solution, and the solution in that case is:
Am I doing it correct if I say that, for the equation to have a solution, and the solution in that case is:
Yes, the condition for a solution to exist is (I carelessly wrote in my previous comment—now corrected—where it should have been .) But you should check your answer to the quadratic equation again. I get .