What is the answer to solvefor x,z=(x+4i)/(1-xi) ?

The answer to solvefor x,z=(x+4i)/(1-xi) is x=(z-(z(4+z^2))/(z^2+1))-(4+z^2)/(z^2+1)i

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solvefor x,z=(x+4i)/(1-xi)

solve for

x, z = \frac{x + 4 i}{1 - x i}

x = (z - \frac{z ( 4 + z ^{2} )}{z ^{2} + 1}) - \frac{4 + z ^{2}}{z ^{2} + 1} i

Solution steps

z = \frac{x + 4 i}{1 - x i}

Substitute

x = a + bi

z = \frac{a + bi + 4 i}{1 - ( a + bi ) i}

Multiply both sides by

1 - (a + bi) i

z (1 - (a + bi) i) = \frac{a + bi + 4 i}{1 - ( a + bi ) i} (1 - (a + bi) i)

Expand

z + z b - i z a = a + i (4 + b)

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

[z + z b = a - z a = 4 + b]

[z + z b = a - z a = 4 + b] : a = z - \frac{z ( 4 + z ^{2} )}{z ^{2} + 1}, b = - \frac{4 + z ^{2}}{z ^{2} + 1}, z \neq = i, z \neq = - i

Substitute back

x = a + bi

x = (z - \frac{z ( 4 + z ^{2} )}{z ^{2} + 1}) - \frac{4 + z ^{2}}{z ^{2} + 1} i

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

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

- \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

What is the answer to solvefor x,z=(x+4i)/(1-xi) ?
The answer to solvefor x,z=(x+4i)/(1-xi) is x=(z-(z(4+z^2))/(z^2+1))-(4+z^2)/(z^2+1)i