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.

Your idea makes sense. The equation has one solution if and only if the discriminant $b^2-4ac$ is zero.
$$b^2-4ac=(-2m)^2-4(m-2)(1+i)=4m^2-(4+4i)m+(8+8i)$$
This is a quadratic equation in $m$ which has the two solutions $2i$ and $1-i$.
