Question

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Answer 1

14)

a) $\dfrac{5-2i}{3+4i}=\dfrac{(5-2i)(3-4i)}{(3+4i)(3-4i)}=\dfrac{15-20i-12i+8i^2}{3^2-(4i)^2}$

$\dfrac{5-2i}{3+4i}=\dfrac{15-32i-8}{9-16i^2}=\dfrac{7-32i}{9+16}=\dfrac{7-32i}{25}$


b) $\dfrac{3+5i}{4-2i}=\dfrac{(3+5i)(4+2i)}{(4-2i)(4+2i)}=\dfrac{12+6i+20i+10i^2}{4^2-(2i)^2}$

$\dfrac{3+5i}{4-2i}=\dfrac{12+26i-10}{16-4i^2}=\dfrac{2+26i}{16+4}=\dfrac{2+26i}{20}=\dfrac{1+13i}{10}$

c) $\dfrac{-2+3i}{3+i}=\dfrac{(-2+3i)(3-i)}{(3+i)(3-i)}=\dfrac{-6+2i+9i-3i^2}{3^2-i^2}$

$\dfrac{-2+3i}{3+i}=\dfrac{-6+11i+3}{9+1}=\dfrac{-3+11i}{10}$