What is the answer to 4z^2-4z+3-2sqrt(3)i=0 ?

The answer to 4z^2-4z+3-2sqrt(3)i=0 is z=-(sqrt(3))/2 i,z=1+(sqrt(3))/2 i

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4z^2-4z+3-2sqrt(3)i=0

4 z^{2} - 4 z + 3 - 23 i = 0

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

Solution steps

4 z^{2} - 4 z + 3 - 23 i = 0

Substitute

z = x + y i

4 (x + y i)^{2} - 4 (x + y i) + 3 - 23 i = 0

Expand

4 (x + y i)^{2} - 4 (x + y i) + 3 - 23 i : (4 x^{2} - 4 y^{2} - 4 x + 3) + i (- 4 y - 23 + 8 x y)

(4 x^{2} - 4 y^{2} - 4 x + 3) + i (- 4 y - 23 + 8 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(4 x^{2} - 4 y^{2} - 4 x + 3) + i (- 4 y - 23 + 8 x y) = 0 + 0 i

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

[4 x^{2} - 4 y^{2} - 4 x + 3 = 0 - 4 y - 23 + 8 x y = 0]

[4 x^{2} - 4 y^{2} - 4 x + 3 = 0 - 4 y - 23 + 8 x y = 0] : (x = 0, x = 1, y = - \frac{3}{2} y = \frac{3}{2})

Substitute back

z = x + y i

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

(x + y i) (2 + 3 i) = 1 - 5 i

4 a - 3 i = 6 + bi

∣ a - i ∣ = 37

x^{2} - 2 i x - 2 = 0

(x + y i) + 4 (1 - i) = 5 - 7 i

What is the answer to 4z^2-4z+3-2sqrt(3)i=0 ?
The answer to 4z^2-4z+3-2sqrt(3)i=0 is z=-(sqrt(3))/2 i,z=1+(sqrt(3))/2 i