Cho số phức z = a + bi a b thuộc R thỏa mãn 2 + i z + 1 - i - 2 - 3i z + i = 2 + 5i. Tính S = 2b – 3b.

Đáp án: 5

Ta có z=a+bi(a,b∈ℝ)⇒z¯=a−bi

Vậy (2+i)(z¯+1−i)−(2−3i)(z+i)=2+5i⇔(2+i)(a−bi+1−i)−(2−3i)(a+bi+i)=2+5i

⇔(2+i)(a+1−(b+1)i)−(2−3i)(a+(b+1)i)=2+5i⇔2(a+1)−2( b+1)i+(a+1)i−(b+1)i2−2a+2( b+1)i−3a−3( b+1)i2=2+5i⇔2a+2−(2b+2)i+(a+1)i+b+1−2a−(2b+2)i+3ai−3b−3=2+5i⇔−2b+(4a−4b−3)i=2+5i⇔−2b=24a−4b−3=5⇔a=1b=−1

Suy ra S=2a−3 b=2⋅1−3⋅(−1)=5.