What is the answer to (1+i)(a+bi)-2(a-bi)=-11+25i ?

The answer to (1+i)(a+bi)-2(a-bi)=-11+25i is a=4,b=7

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(1+i)(a+bi)-2(a-bi)=-11+25i

(1 + i) (a + bi) - 2 (a - bi) = - 11 + 25 i

a = 4, b = 7

Solution steps

(1 + i) (a + bi) - 2 (a - bi) = - 11 + 25 i

Expand

(1 + i) (a + bi) - 2 (a - bi) : (- a - b) + i (a + 3 b)

(- a - b) + i (a + 3 b) = - 11 + 25 i

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

[- a - b = - 11 a + 3 b = 25]

[- a - b = - 11 a + 3 b = 25] : a = 4, b = 7

a = 4, b = 7

(1 + i) (a + 2 b) - (3 - 2 i) (a - b) = 8 + 3 i

- 4 x^{2} + 4 y^{2} i = 8 x y, z = x + i y

x^{2} - (- 2) = 7 i

i (x) = \frac{1 x ^{2} + 1}{x - 4}

(x + i y - 1)^{2} = (x + i y - i)^{2}

What is the answer to (1+i)(a+bi)-2(a-bi)=-11+25i ?
The answer to (1+i)(a+bi)-2(a-bi)=-11+25i is a=4,b=7