Answer
$a)$ $\dfrac{41}{25}-\dfrac{12}{25}i$
$b)$ $-20+0i$
Work Step by Step
$a)$ $\dfrac{8+3i}{4+3i}$
Multiply the numerator and the denominator of this expression by the complex conjugate of the denominator:
$\dfrac{8+3i}{4+3i}=\dfrac{8+3i}{4+3i}\cdot\dfrac{4-3i}{4-3i}=\dfrac{(8+3i)(4-3i)}{4^{2}-(3i)^{2}}=...$
$...=\dfrac{32-24i+12i-9i^{2}}{16-9i^{2}}=...$
Substitute $i^{2}$ by $-1$ and simplify:
$...=\dfrac{32-24i+12i-9(-1)}{16-9(-1)}=\dfrac{32-12i+9}{16+9}=\dfrac{41-12i}{25}=...$
$...=\dfrac{41}{25}-\dfrac{12}{25}i$
$b)$ $\sqrt{-10}\cdot\sqrt{-40}$
Evaluate the product and simplify:
$\sqrt{-10}\cdot\sqrt{-40}=i\sqrt{(10)}.i\sqrt{(40)}=i^2\sqrt{400}= -1. \sqrt{400} = -20+0i$
