Logarithm of imaginary numbers?

Question

How do I solve this question?

I tried using the quadratic formula on the question equation and got

$x_1 = 0.25 +1.089724..i = \ln r$

$x_2 = 0.25 -1.089724..i = \ln s$

I know $\ln x = \log_ex$, but how do I complete the questions with the imaginary numbers?

Answer 1

Actually the key of solving this question is to make use of the RELATION OF ROOTS.

Recall:

If $\forall a,b,c\in\Bbb{R}$, given quadratic of $x$, $ax^2+bx+c=0$,

which can be converted to the form of $x^2+\frac bax+\frac ca=0$,

we say that $-\frac ba$ is the sum of roots and $\frac ca$ is the product of roots.

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In this case, $\ln r$ and $\ln s$ are roots, so you can make use of these relations.

Answer 2

For the $a)$ part, use Vieta's Formula: Sum of roots of a quadratic equation is $\frac{-b}a$. So answer would be: $\frac12$.
For $b)$ $\frac54$, using Vieta's Formulas for product of roots: $\frac ca$.
For $c)$ the expression becomes $$\frac{\ln r+\ln s}{\ln r\ln s}=\frac{\frac12}{\frac54}=\frac25$$ For $d)$ use the properties of logarithms that $\log_rs=\frac{\log s}{\log r}$ and $a^2+b^2=(a+b)^2-2ab$.
So the given expression becomes $$\frac{(\log r+\log s)^2-2\log r\log s}{\log r.\log s}=\frac{\frac14-\frac{10}4}{\frac54}=\frac{-9}{5}$$