What is the answer to 0=z^2+iz-6 ?

The answer to 0=z^2+iz-6 is z=(sqrt(23))/2-1/2 i,z=-(sqrt(23))/2-1/2 i

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0=z^2+iz-6

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

z = \frac{23}{2} - \frac{1}{2} i, z = - \frac{23}{2} - \frac{1}{2} i

Solution steps

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

Substitute

z = x + y i

0 = (x + y i)^{2} + i (x + y i) - 6

Expand

(x + y i)^{2} + i (x + y i) - 6 : (x^{2} - y^{2} - y - 6) + i (x + 2 x y)

0 = (x^{2} - y^{2} - y - 6) + i (x + 2 x y)

Rewrite

0

in standard complex form:

0 + 0 i

0 + 0 i = (x^{2} - y^{2} - y - 6) + i (x + 2 x y)

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

[0 = x^{2} - y^{2} - y - 6 0 = x + 2 x y]

[0 = x^{2} - y^{2} - y - 6 0 = x + 2 x y] : (x = \frac{23}{2}, x = - \frac{23}{2}, y = - \frac{1}{2} y = - \frac{1}{2})

Substitute back

z = x + y i

z = \frac{23}{2} - \frac{1}{2} i, z = - \frac{23}{2} - \frac{1}{2} i

(7 + m) + (4 l - 10) i = 3 - 6 i

(- 2 - 6 i) - p (4 - 2 i) = q i

(x - y i) + (y + x i) = (5 + i)

x, g (3 - i) = x^{2} - 6 x

(x + 4) i - (y - 3) j = 2 y (i + j)

What is the answer to 0=z^2+iz-6 ?
The answer to 0=z^2+iz-6 is z=(sqrt(23))/2-1/2 i,z=-(sqrt(23))/2-1/2 i