Ans: (a) $\dfrac{1}{5}-\dfrac{2}{5}i$ (b) (i) $p=-4$, $q=20$ (ii) $r\ge 16$
-
$\begin{array}{rcl}
\dfrac{1}{1+2i} & = & \dfrac{1}{1+2i} \times \dfrac{1-2i}{1-2i} \\
& = & \dfrac{1-2i}{1-(2i)^2} \\
& = & \dfrac{1-2i}{1+4} \\
& = & \dfrac{1}{5}-\dfrac{2}{5}i
\end{array}$ -
- Consider the sum of the roots,
$\begin{array}{rcl}
-p & = & \dfrac{10}{1+2i}+\dfrac{10}{1-2i} \\
p & = & -\dfrac{10(1-2i)+10(1+2i)}{(1+2i)(1-2i)} \\
& = & -\dfrac{10-20i+10+20i}{5} \\
& = & -4
\end{array}$Consider the product of the roots,
$\begin{array}{rcl}
q & = & \dfrac{10}{1+2i}\times\dfrac{10}{1-2i} \\
& = & \dfrac{100}{(1+2i)(1-2i)} \\
& = & \dfrac{100}{5} \\
& = & 20
\end{array}$ - By the result of (b)(i), we have
$\begin{array}{rcl}
x^2-4p+20 & = & r \\
x^2-4p+(20-r) & = & 0
\end{array}$For the equation has real roots,
$\begin{array}{rcl}
\Delta & \ge & 0 \\
(-4)^2-4(1)(20-r) & \ge & 0 \\
16 -80+4r & \ge & 0 \\
4r & \ge & 64 \\
r & \ge & 16
\end{array}$
- Consider the sum of the roots,
