Logarithm of imaginary numbers?

By the quadratic formula the roots of the given equation are

$x = \dfrac{2 \pm \sqrt{(-2)^{2} - 4*4*5}}{2*4} = \dfrac{1 \pm i*\sqrt{19}}{4}.$

Without loss of generality we can assign $\ln(r)$ and $\ln(s)$ as the positive and negative roots, respectively.

Now by the change-of-base rule we have that

$\log_{r}(s) + \log_{s}(r) = \dfrac{\ln(s)}{\ln(r)} + \dfrac{\ln(r)}{\ln(s)}.$

Next, we find that

$\dfrac{\ln(r)}{\ln(s)} = \dfrac{1 + i*\sqrt{19}}{1 - i*\sqrt{19}} \times \dfrac{1 + i*\sqrt{19}}{1 + i*\sqrt{19}} = \dfrac{1 - 19 + i*2\sqrt{19}}{20} = \dfrac{-9 + i*\sqrt{19}}{10},$

and so

$\dfrac{\ln(r)}{\ln(s)} + \dfrac{\ln(s)}{\ln(r)} = \dfrac{-9 + i*\sqrt{19}}{10} + \dfrac{10}{-9 + i*\sqrt{19}} = \dfrac{81 - 19 - i*18\sqrt{19}}{10*(-9 + i*\sqrt{19})} =$

$\dfrac{162 - i*18\sqrt{19}}{10*(-9 + i*\sqrt{19})} = \dfrac{-18*(-9 + i*\sqrt{19})}{10*(-9 + i*\sqrt{19})} = -\dfrac{18}{10} = \boxed{-1.8}.$

Let $ln r=a$ and $ln s=b$ . This means $log_{r}s+log_{s}r$ $=\frac{ln s}{ln r}+ \frac{ln s}{ln r}$ $=\frac{a}{b} +\frac{b}{a}$ $=\frac{a^{2}+b^{2}}{ab}$ $=\frac{(a+b)^{2}-2ab}{ab}$ $=\frac{(\frac{1}{2})^{2}-2 \cdot \frac{5}{4}}{\frac{5}{4}}$ $=-1.8$