Question

AJUDA URGENTE! PFVR
Considere a=8+3i e b=6+2i. Determine: a•b e a/b

Answer 1

a = 8 +3.i

b = 6 +2.i

 a.b

( 8 +3.i).( 6 +2.i)

8.6 +8.2.i +3.i.6 +3.i.2.i

48 +16.i +18.i +6.i²

48 +34.i +6.(-1)

48 -6 +34.i

42 +34.i

 a/b

$\frac{8+3.i}{6+2.i}\\\frac{8+3.i}{6+2.i}.\frac{(6-2.i)}{(6-2.i)}\\\frac{48-16.i+18.i-6.i^2}{36-4.i^2}\\\frac{48+6+2.i}{36+4}\\\frac{54+2.i}{40}\\\frac{2.(27+i)}{40}\\\frac{27+i}{20}\\\frac{27}{20}+\frac{i}{20}$

Dúvidas só perguntar!

Answer 2

a=8+3i

b=6+2i

$ab=(8+3i)(6+2i) \\ ab=48+16i+18i+6{i}^{2} \\ ab=48+34i+6.(-1) \\ ab=48+34i-6 \\ ab=42+34i$

$\frac{a}{b}=\frac{8+3i}{6+2i} \\ \frac{a}{b}=\frac{(8+3i)}{(6+2i)}\times\frac{(6-2i)}{(6-2i)}$

$\frac{a}{b}=\frac{48+2i-6{i}^{2}}{{6}^{2}-{(2i)}^{2}}$

$\frac{a}{b} =\frac{48+2i-6.(-1)}{36-4{i}^{2}}$

$\frac{a}{b}=\frac{48+2i+6}{36-4.(-1)}$

$\frac{a}{b} =\frac{54+2i}{36+4}$

$\frac{a}{b}=\frac{54}{40}+\frac{2}{40}i$

$\frac{a}{b}=\frac{27}{20}+\frac{1}{20}i$