Quadratic equation: The relationship between solutions and coefficients 2

Question:
The equation, $kx^2 -(k+3)x - 1 =0$, which has real coefficients, has complex roots, $a+ib$ and $a-ib$. In order to have those roots, find the range of $k$. If they are pure imaginary roots, find the value of $k$.

Answer:
The discriminant of the quadratic equation, $ax^2+bx+c=0$, is
\[
D = b^2 - 4ac
\]
For complex roots, we must have $D<0$ and $k \neq 0$. Thus,
\begin{eqnarray*}
D &=& (k+3)^2 +4k    \\
    &=& k^2+10k+9      \\
    &=& (k+1)(k+9)  < 0
\end{eqnarray*}
The range of $k$ is
\[
-9 < k < -1 \quad \mathrm{but} \quad k \neq 0
\]
If the roots are pure imaginary, we let the roots be $ib$ and $-ib$. We have the following relationship:
\[
ib + (-ib) = - \frac{-(k+3)}{k}
\]
This is from the relationship between the roots and coefficients. Then,
\begin{eqnarray*}
0 &=& k +3  \\
\rightarrow k &=& -3
\end{eqnarray*}
This satisfies $-9 < k < -1$.

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