What is the answer to (1+i)(a+2b)-(3-2i)(a-b)=8+3i ?

The answer to (1+i)(a+2b)-(3-2i)(a-b)=8+3i is a=1,b=2

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

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

a = 1, b = 2

Solution steps

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

Rewrite

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

in standard complex form:

(- 2 a + 5 b) + 3 ai

(- 2 a + 5 b) + 3 ai = 8 + 3 i

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

[- 2 a + 5 b = 8 3 a = 3]

[- 2 a + 5 b = 8 3 a = 3] : a = 1, b = 2

a = 1, b = 2

- 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}

(7 a + bi) + (- 3 b + ai) = (2 + 5 i)

What is the answer to (1+i)(a+2b)-(3-2i)(a-b)=8+3i ?
The answer to (1+i)(a+2b)-(3-2i)(a-b)=8+3i is a=1,b=2