What is the answer to -2(a+bi)+i(a-bi)=7i-8 ?

The answer to -2(a+bi)+i(a-bi)=7i-8 is a=3,b=-2

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-2(a+bi)+i(a-bi)=7i-8

- 2 (a + bi) + i (a - bi) = 7 i - 8

a = 3, b = - 2

Solution steps

- 2 (a + bi) + i (a - bi) = 7 i - 8

Expand

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

(- 2 a + b) + i (a - 2 b) = 7 i - 8

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

[- 2 a + b = - 8 a - 2 b = 7]

[- 2 a + b = - 8 a - 2 b = 7] : a = 3, b = - 2

a = 3, b = - 2

x^{2} + x + 3 i = 0

\frac{1}{0} = a + bi

(x^{2} - 2 x - 15) + (3 y - 9) i = 3 i

3 ai - b - 3 i = a - 5 + bi

∣ z - 1 + 3 i ∣ = 2

What is the answer to -2(a+bi)+i(a-bi)=7i-8 ?
The answer to -2(a+bi)+i(a-bi)=7i-8 is a=3,b=-2