Question

If (x – 2) + iy = 3 + 4i then x + y is

Answer 1

Answer:

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Answer 2

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(x + yi)^2 = x^2 + 2xyi - y^2 = (x^2 - y^2) + 2xyi = 3 + 4i

On this basis we set up a system of equations as follows:

x^2 - y^2 = 3

2xyi = 4i — > xy = 2 — > y = 2/x

x^2 - 4/x^2 = 3 [substitute 2/x for y in the first equation]

x^4 - 4 = 3x^2

x^4 - 3x^2 - 4 = 0

(x^2 - 4) (x^2 + 1) = 0 [factor, and note the two differences of squares]

(x + 2) (x - 2) (x + i) (x - i) = 0

So the roots of x are { -2, 2, -i, i }. Plug these values into the equation y = 2/x to yield the corresponding values of y: { -1, 1, 2i, -2i }. So the solution set, in terms of ordered pairs (x, y), is {(x, y)} = { (-2, -1), (2, 1), (-i, 2i), (i, - 2i) }