What is the answer to 6/(x+yi)+(2-i)^2=4-3i ?

The answer to 6/(x+yi)+(2-i)^2=4-3i is x=3,y=-3

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6/(x+yi)+(2-i)^2=4-3i

\frac{6}{x + y i} + (2 - i)^{2} = 4 - 3 i

x = 3, y = - 3

Solution steps

\frac{6}{x + y i} + (2 - i)^{2} = 4 - 3 i

Multiply both sides by

x + y i

\frac{6}{x + y i} (x + y i) + (2 - i)^{2} (x + y i) = 4 (x + y i) - 3 i (x + y i)

Expand

(6 + 3 x + 4 y) + i (3 y - 4 x) = (4 x + 3 y) + i (- 3 x + 4 y)

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

[6 + 3 x + 4 y = 4 x + 3 y 3 y - 4 x = - 3 x + 4 y]

[6 + 3 x + 4 y = 4 x + 3 y 3 y - 4 x = - 3 x + 4 y] : x = 3, y = - 3

x = 3, y = - 3

z^{2} - (1 + 4 i) z + (3 + 11 i) = 0

x + i y = 2 i - 7

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

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

x - 2 i + 3 y i = - 5

What is the answer to 6/(x+yi)+(2-i)^2=4-3i ?
The answer to 6/(x+yi)+(2-i)^2=4-3i is x=3,y=-3