What is the answer to 2(a+ib)+(3+i)(a-ib)=4-i ?

The answer to 2(a+ib)+(3+i)(a-ib)=4-i is a= 1/2 ,b= 3/2

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2(a+ib)+(3+i)(a-ib)=4-i

2 (a + ib) + (3 + i) (a - ib) = 4 - i

a = \frac{1}{2}, b = \frac{3}{2}

Solution steps

2 (a + ib) + (3 + i) (a - ib) = 4 - i

Expand

2 (a + ib) + (3 + i) (a - ib) : (5 a + b) + i (a - b)

(5 a + b) + i (a - b) = 4 - i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[5 a + b = 4 a - b = - 1]

[5 a + b = 4 a - b = - 1] : a = \frac{1}{2}, b = \frac{3}{2}

a = \frac{1}{2}, b = \frac{3}{2}

1 = ai + b

(1 + 2 i) a + (3 - 5 i) b = 1 - 3 i

7 x + x i - 3 y i = 3 y + 9

(x + y i) (2 - 4 i) = 14 - 18 i

3 z^{2} - 2 + z - 3 i z^{2} + 2 i + z i = 0

What is the answer to 2(a+ib)+(3+i)(a-ib)=4-i ?
The answer to 2(a+ib)+(3+i)(a-ib)=4-i is a= 1/2 ,b= 3/2