What is the answer to 1+3iez^2=(1-i) ?

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

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1+3iez^2=(1-i)

1 + 3 i e z^{2} = (1 - i)

z = \frac{1}{3 e} i, z = - \frac{1}{3 e} i

Solution steps

1 + 3 i e z^{2} = (1 - i)

Substitute

z = x + y i

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

Expand

1 + 3 i e (x + y i)^{2} : (1 - 6 e x y) + i (3 e x^{2} - 3 e y^{2})

(1 - 6 e x y) + i (3 e x^{2} - 3 e y^{2}) = 1 - i

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

[1 - 6 e x y = 1 3 e x^{2} - 3 e y^{2} = - 1]

[1 - 6 e x y = 1 3 e x^{2} - 3 e y^{2} = - 1] : x = 0, x = 0, y = \frac{1}{3 e} y = - \frac{1}{3 e}

Substitute back

z = x + y i

z = \frac{1}{3 e} i, z = - \frac{1}{3 e} i

x + (y - 2) i = 3

x^{2} + (2 - 2 i) x - (4 + 2 x) = 0

2 x - 20 i = 8 - (4 y) i

(x + i y) (1 + i) = 3 - i

aa - 2 a - 3 i = 0

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