What is the answer to solvefor z,2z^2-2iz-1=0 ?

The answer to solvefor z,2z^2-2iz-1=0 is z= 1/2+1/2 i,z=-1/2+1/2 i

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solvefor z,2z^2-2iz-1=0

solve for

z, 2 z^{2} - 2 i z - 1 = 0

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

Solution steps

2 z^{2} - 2 i z - 1 = 0

Substitute

z = x + y i

2 (x + y i)^{2} - 2 i (x + y i) - 1 = 0

Expand

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

(2 x^{2} - 2 y^{2} + 2 y - 1) + i (- 2 x + 4 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(2 x^{2} - 2 y^{2} + 2 y - 1) + i (- 2 x + 4 x y) = 0 + 0 i

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

[2 x^{2} - 2 y^{2} + 2 y - 1 = 0 - 2 x + 4 x y = 0]

[2 x^{2} - 2 y^{2} + 2 y - 1 = 0 - 2 x + 4 x y = 0] : (x = \frac{1}{2}, x = - \frac{1}{2}, y = \frac{1}{2} y = \frac{1}{2})

Substitute back

z = x + y i

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

\frac{1 - 2 i}{4 + 2 i} = a + bi

x i n^{0} = 3 y^{2} + x - 1

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

z^{2} - 2 z + 9 + 6 i = 0

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

What is the answer to solvefor z,2z^2-2iz-1=0 ?
The answer to solvefor z,2z^2-2iz-1=0 is z= 1/2+1/2 i,z=-1/2+1/2 i