complex number equation?

Question

help solving this equation

2z^2 + 2iz -1 =0

Answer 1

a=2
b=2i
c=-1
now just plug and chug!

-b+/-[√(b^2 -4ac)]/2a
-2i +/- [√(-2i)^2 - 4(2)(-1)]/2(2)
-2i +/- [√-4+8]/4

z= [-i +/- 1]/2
simple and quick!

Answer 2

I assume that z is the complex variable x + iy;

The equation expands to

2[x^2 + 2ixy - y^2] + 2ix - 2y - 1 = 0

collect the real terms and imaginary terms

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

For the whole to be zero, both real and imaginary parts must be zero

first the imaginary part: 4xy + 2x = 0, y = -0.5

put this into the real part

2x^2 - 0.5 - +1 - 1 = 0

2x^2 = 0.5, x = √.25 = ±0.5

The answer is then ±0.5 - 0.5i

Answer 3

2z^2 + 2iz -1 =0

By the quadratic formula

z= (-2i +/- sqrt((2i)^2 - 4(2)(-1)))/(2*2)
= (-2i +/- sqrt(4i^2 + 8))/4
= (-2i +/- sqrt(-4 + 8))/4
= (-2i +/- sqrt(4))/4
= (-2i +/- 2)/4
= -i/2 +/- 1/2

z = 1/2 - i/2 or -1/2 - i/2