What is the answer to -2iez^2=-1+i,z^1=3 ?

The answer to -2iez^2=-1+i,z^1=3 is

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-2iez^2=-1+i,z^1=3

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

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

Solution steps

- 2 i e z^{2} = - 1 + i

Substitute

z = x + y i

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

Expand

- 2 i e (x + y i)^{2} : 4 e x y - 2 e i (x^{2} - y^{2})

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

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

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

[4 e x y = - 1 - 2 e x^{2} + 2 e y^{2} = 1] : x = \frac{2 - 1}{2 e}, x = - \frac{2 - 1}{2 e}, y = - \frac{1}{2 e 2 - 1} y = \frac{1}{2 e 2 - 1}

Substitute back

z = x + y i

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

(i) x^{2} - 16 = 0, (ii) 6 x^{2} - 24 = 0

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

(2 i + 1)^{2} = (m + ni) (2 - i)

2 (x - y i) + 5 (x + y i) + (x + y i)^{2} = 0

- 0.008 i = \frac{1}{x \cdot i}