What is the answer to z-(1/z)=3+2i ?

The answer to z-(1/z)=3+2i is z=-(-3+2sqrt(3))/2+(-sqrt(3)+2)/2 i,z=(3+2sqrt(3))/2+(sqrt(3)+2)/2 i

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z-(1/z)=3+2i

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

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

Solution steps

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

Substitute

z = x + y i

x + y i - \frac{1}{x + y i} = 3 + 2 i

Multiply both sides by

x + y i

x (x + y i) + y i (x + y i) - \frac{1}{x + y i} (x + y i) = 3 (x + y i) + 2 i (x + y i)

Expand

(x^{2} - y^{2} - 1) + 2 i x y = (3 x - 2 y) + i (2 x + 3 y)

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

[x^{2} - y^{2} - 1 = 3 x - 2 y 2 x y = 2 x + 3 y]

[x^{2} - y^{2} - 1 = 3 x - 2 y 2 x y = 2 x + 3 y] : (x = - \frac{- 3 + 2 3}{2}, x = \frac{3 + 2 3}{2}, y = \frac{- 3 + 2}{2} y = \frac{3 + 2}{2})

Substitute back

z = x + y i

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

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

(2 - i)^{2} - (2 - y i) = x + 5 i

∣ z - (1 - i) ∣ = 2

a - bi = i (a + bi)

∣ z - 1 + i ∣ = 3

What is the answer to z-(1/z)=3+2i ?
The answer to z-(1/z)=3+2i is z=-(-3+2sqrt(3))/2+(-sqrt(3)+2)/2 i,z=(3+2sqrt(3))/2+(sqrt(3)+2)/2 i