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=-51/25 ,y= 118/25

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

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

x = - \frac{51}{25}, y = \frac{118}{25}

Solution steps

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

Multiply both sides by

x + y i + (2 - i)^{2}

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

Expand

6 = (4 x + 3 y) + i (- 25 - 3 x + 4 y)

Rewrite

6

in standard complex form:

6 + 0 i

6 + 0 i = (4 x + 3 y) + i (- 25 - 3 x + 4 y)

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

[6 = 4 x + 3 y 0 = - 25 - 3 x + 4 y]

[6 = 4 x + 3 y 0 = - 25 - 3 x + 4 y] : x = - \frac{51}{25}, y = \frac{118}{25}

x = - \frac{51}{25}, y = \frac{118}{25}

∣ z - i ∣ = 2 z + i

3 i z^{2} - 2 z = - i

x, s i f = y x^{2} - x y^{3}

\frac{( 18 + i )}{4 + 3 i} = a + bi

in (in (in x)) = y

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=-51/25 ,y= 118/25