What is the answer to 2z^2+(-1+i)z+3+i=0 ?

The answer to 2z^2+(-1+i)z+3+i=0 is z=i,z= 1/2-3/2 i

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2z^2+(-1+i)z+3+i=0

2 z^{2} + (- 1 + i) z + 3 + i = 0

z = i, z = \frac{1}{2} - \frac{3}{2} i

Solution steps

2 z^{2} + (- 1 + i) z + 3 + i = 0

Substitute

z = x + y i

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

Expand

2 (x + y i)^{2} + (- 1 + i) (x + y i) + 3 + i : (2 x^{2} - 2 y^{2} - x - y + 3) + i (1 + x - y + 4 x y)

(2 x^{2} - 2 y^{2} - x - y + 3) + i (1 + x - y + 4 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(2 x^{2} - 2 y^{2} - x - y + 3) + i (1 + x - y + 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} - x - y + 3 = 0 1 + x - y + 4 x y = 0]

[2 x^{2} - 2 y^{2} - x - y + 3 = 0 1 + x - y + 4 x y = 0] : (x = 0, x = \frac{1}{2}, y = 1 y = - \frac{3}{2})

Substitute back

z = x + y i

z = i, z = \frac{1}{2} - \frac{3}{2} i

a + b - 2 abi = 5 - 12 i

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

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

\frac{a + 7 i}{3 - 2 i} = 1 + bi

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

What is the answer to 2z^2+(-1+i)z+3+i=0 ?
The answer to 2z^2+(-1+i)z+3+i=0 is z=i,z= 1/2-3/2 i