What is the answer to roots z^2+2z-isqrt(3) ?

The answer to roots z^2+2z-isqrt(3) is z=(sqrt(6)-2)/2+1/(sqrt(2))i,z=-(sqrt(6)+2)/2-1/(sqrt(2))i

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roots z^2+2z-isqrt(3)

roots

z^{2} + 2 z - i 3

z = \frac{6 - 2}{2} + \frac{1}{2} i, z = - \frac{6 + 2}{2} - \frac{1}{2} i

Solution steps

z^{2} + 2 z - i 3

The roots are the intercepts with the x-axis

(y = 0)

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

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

[x^{2} - y^{2} + 2 x = 0 - 3 + 2 y + 2 x y = 0] : (x = \frac{6 - 2}{2}, x = - \frac{6 + 2}{2}, y = \frac{1}{2} y = - \frac{1}{2})

Substitute back

z = x + y i

z = \frac{6 - 2}{2} + \frac{1}{2} i, z = - \frac{6 + 2}{2} - \frac{1}{2} i

\frac{s}{( x + 5 ) ( x + 2 ) ( x + 4 )}

\frac{1}{5 ( 5 - x )}

x^{2 (14 - k)}

\frac{4}{5} (\frac{4}{5} ln (22950) + \frac{1}{5} ln (4006000))

\frac{e ^{- 11 s}}{12 s ( 100 s + 1 )}

What is the answer to roots z^2+2z-isqrt(3) ?
The answer to roots z^2+2z-isqrt(3) is z=(sqrt(6)-2)/2+1/(sqrt(2))i,z=-(sqrt(6)+2)/2-1/(sqrt(2))i