What is the answer to solvefor c,z=(-3+3i)/(2-c) ?

The answer to solvefor c,z=(-3+3i)/(2-c) is c=-(-3-2z)/z-3/z i

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solvefor c,z=(-3+3i)/(2-c)

solve for

c, z = \frac{- 3 + 3 i}{2 - c}

c = - \frac{- 3 - 2 z}{z} - \frac{3}{z} i

Solution steps

z = \frac{- 3 + 3 i}{2 - c}

Substitute

c = x + y i

z = \frac{- 3 + 3 i}{2 - ( x + y i )}

Multiply both sides by

2 - x - i y

z (2 - x - i y) = \frac{- 3 + 3 i}{2 - ( x + y i )} (2 - x - i y)

Expand

2 z - z x - i zy = - 3 + 3 i

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

[2 z - z x = - 3 - zy = 3]

[2 z - z x = - 3 - zy = 3] : x = - \frac{- 3 - 2 z}{z}, y = - \frac{3}{z}, z \neq = 0

Substitute back

c = x + y i

c = - \frac{- 3 - 2 z}{z} - \frac{3}{z} i

z^{2} = - 15 + 8 i

\frac{( - 6 i + n - 3 n + 5 )}{4 + n ^{2}} = 0

(a + 2 i) (1 + 2 i) = 4 - bi

z^{2} - \frac{2 i + 1}{2 - i} = 0

∣ x + y i ∣ = y - i x

What is the answer to solvefor c,z=(-3+3i)/(2-c) ?
The answer to solvefor c,z=(-3+3i)/(2-c) is c=-(-3-2z)/z-3/z i