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

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

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(a+bi)=(1+i)*(a-bi)+3-2i

(a + bi) = (1 + i) \cdot (a - bi) + 3 - 2 i

a = - 4, b = - 3

Solution steps

(a + bi) = (1 + i) (a - bi) + 3 - 2 i

Expand

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

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

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

[a = a + b + 3 b = - 2 + a - b] : a = - 4, b = - 3

a = - 4, b = - 3

- \frac{1}{z - 2 i} + 1 - z = 0

(ab + 6) + (- 2 a + 3 b) i = 12

\frac{3 - x i}{1 + 2 i} = y + 2 i

\frac{2 + ai}{6 + bi} = 1 + 5 i

∣ z - 2 i ∣ = 5

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