Question

Efetue (1+i)² . (3/1-1) + (1-i/1+i)




a) 1 - 3i
b) +3i
c) -3i
d) -1
e) -1 + 3i

Answer 1

R: Letra E

veja

$(1+i)^2.( \frac{3}{1-i} )+( \frac{1-i}{1+i} )=$

vamos calcular separado, fica mais fácil

$(1+i)^2=1+2i+i^2=1+2i+(-1)=1+2i-1=2i$

e

$\frac{3}{1-i} + \frac{1-i}{1+i} = \frac{3(1+i)+(1-i)^2}{(1+i)(1-i)} = \frac{3+3i+1-2i+i^2}{1^2-i^2} =$

$\frac{3+3i+1-2i+(-1)}{1-(-1)} = \frac{4+i-1}{1+1} = \frac{3+i}{2}$

vamos multiplicar

$2i( \frac{3+i}{2} )= \frac{6i+2i^2}{2} = \frac{6i+2(-1)}{2} = \frac{6i-2}{2} =$

$\frac{2(3i-1)}{2} =3i-1$  ou como está  -1+3i

Answer 2

$(1 + i)^2 * (\dfrac{3i}{1-i}) + ( \dfrac{1-i}{1+i} ) \\ \\ \\ (1 + i) * (1 + i) * (\dfrac{3 + 3i}{2}) - i) \\ \\ \\ 2i * (\dfrac{3 + 3i}{2}) - i) \\ \\ \\ 2i * (\dfrac{3 + i}{2})) \\ \\ \\ => -1 + 3i$