So I came across this quadratic, $z^2(1-2i)+6iz+-2i-1$, and used the quadratic equation to get its roots, $z_1 = \frac{2-i}{5}, z_2 = 2-i$. However, when I used this in a limit it came out differently.
$$\lim_{z \to \frac{2-i}{5}} (z – \frac{2-i}{5})(\frac{2}{z^2(1-2i)+6iz+-2i-1)} = \frac{-i}{2}$$
$$\lim_{z \to \frac{2-i}{5}} (z – \frac{2-i}{5})(\frac{2}{(z – \frac{2-i}{5})(z – 2 + i)}) = -1 -\frac{-i}{2}$$
This has lead me to believe the quadratic may have three roots. Not sure though, just can't see why else they wouldn't be equal – hoping someone can shed some light.
Thanks, Jack
Best Answer
If $\,az^2+bz+c=0\,$ has roots $\,z_1\,$ and $\,z_2\,$, then $\, az^2+bz+c=a(z−z_1)(z−z_2)\,$ factors. Notice carefully the need to include the factor $\,a\,$ which you forgot about and this is your mistake.