What is the answer to (8+i)(x+yi)=(8-i)(x-yi) ?

The answer to (8+i)(x+yi)=(8-i)(x-yi) is x=-8y,y=y

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(8+i)(x+yi)=(8-i)(x-yi)

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

x = - 8 y, y = y

Solution steps

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

Expand

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

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

[8 x - y = 8 x - y x + 8 y = - x - 8 y]

[8 x - y = 8 x - y x + 8 y = - x - 8 y] : x = - 8 y, y = y

x = - 8 y, y = y

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

x - 2 i + 3 y i = - 5

∣ z + i ∣ = 3

(3 x - i) (2 + i) + (x - i y) (1 + 2 i) = 5 + 6 i

\frac{( 3 + 2 i )}{z ^{2}} = 1

What is the answer to (8+i)(x+yi)=(8-i)(x-yi) ?
The answer to (8+i)(x+yi)=(8-i)(x-yi) is x=-8y,y=y