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Efetue: (números complexos)
a)(-3i+4).(-2+5i)
b)(1+i).(2-i).(3+2i)


RESOLVENDO

a)(-3i+4).(-2+5i) = 6i - 15. i^2 - 8 + 20i ==> 15 - 8 +6i + 20i ==> 7 + 26i
b)(1+i).(2-i).(3+2i)
(1+i)(2-i)= 2 - i + 2i - i^2 ==> 2 + 1 - i + 2i ==> 3 + i
( 3 + i)(3 + 2i)= 9 + 6i +3i + 2i^2 = 9 - 2 + 6i + 3i ==> 7 + 9i

A)
\( (-3i+4)\cdot(-2+5i)=\\\\ +6i-15i^2-8+20i =\\\\ \boxed{i^2 = -1}\\\\ 6i-(15\cdot-1)-8+20i=\\\\ \boxed{26i+7} \)
b)
\( (1+i)\cdot(2-i)\cdot(3+2i) = \\\\ (1+i)\cdot(2-i) = 2-i+2i-i^2=\\\\ \boxed{(3+i)}\\\\ (3+i)\cdot(3+2i)\\\\ 9+6i+3i+2i^2 = \boxed{7+9i} \)



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