finding a solution for $m$ given $(1+i)z^2-2mz+m-2=0$ Given the equation: $(1+i)z^2-2mz+m-2=0$, while $z$ is complex and $m$ is a parameter.
For which values of $m$ the equation has one solution?
So my idea was to use: $b^2-4ac=0$ for $ax^2+bx+c=0$
But it leads to difficult computation which i could not solve.
Is there any other way? or any way to solve this question?
Thanks.

You should write your equation as:
$(1+i)\cdot z^2 + (-2 m -1)\cdot z + m$
now you can use your idea:
$b^2-4ac$
in your case:
$(-2 m-1)^2-4(1+i)(m) = 4m^2 + 4m + 1 - 4m - 4mi = 4m^2 - 4mi + 1 = 0$
Solve this formula for quadratic equation leads to $ m_{1/2} = \frac{4i\pm\sqrt{-32}}{8} = \frac{4i\pm4\sqrt{2}i}{8} = \frac{1}{2}i \pm \frac{\sqrt{2}}{2}i$
