What is the answer to (a-bi)(a+bi)+(a-bi)=3-i ?

The answer to (a-bi)(a+bi)+(a-bi)=3-i is

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(a-bi)(a+bi)+(a-bi)=3-i

(a - bi) (a + bi) + (a - bi) = 3 - i

(a = 1, a = - 2, b = 1 b = 1)

Solution steps

(a - bi) (a + bi) + (a - bi) = 3 - i

Simplify

(a - bi) (a + bi) + (a - bi) : (a^{2} + b^{2} + a) - ib

(a^{2} + b^{2} + a) - ib = 3 - i

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

[a^{2} + b^{2} + a = 3 - b = - 1]

[a^{2} + b^{2} + a = 3 - b = - 1] : (a = 1, a = - 2, b = 1 b = 1)

(a = 1, a = - 2, b = 1 b = 1)

(x - y) + 8 i = 6 + y i

2 p - q + i (7 i + 3) = 2 (2 i - q) - i (p + 3 q)

y - 5 = - i [x - (- 5)]

(2 + 3 i) (x - y) i = 2 - 10 i

(6 - l) + (3 m) i = - 12 + 27 i

What is the answer to (a-bi)(a+bi)+(a-bi)=3-i ?
The answer to (a-bi)(a+bi)+(a-bi)=3-i is