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

The answer to iz^2-z+i=0 is z=-(1+sqrt(5))/2 i,z=(sqrt(5)-1)/2 i

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iz^2-z+i=0

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

z = - \frac{1 + 5}{2} i, z = \frac{5 - 1}{2} i

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

[- 2 x y - x = 0 1 + x^{2} - y - y^{2} = 0]

[- 2 x y - x = 0 1 + x^{2} - y - y^{2} = 0] : (x = 0, x = 0, y = - \frac{1 + 5}{2} y = \frac{5 - 1}{2})

Substitute back

z = x + y i

z = - \frac{1 + 5}{2} i, z = \frac{5 - 1}{2} i

z^{2} - 2 z + 1 - 2 i = 0

z^{2} - 2 z - 4 i = 2

\frac{4 + 3 i}{m + ni} = 1 + 2 i

5 + 8 i = x + y i

\frac{1 + ai}{7 + bi} = 3 + 5 i

What is the answer to iz^2-z+i=0 ?
The answer to iz^2-z+i=0 is z=-(1+sqrt(5))/2 i,z=(sqrt(5)-1)/2 i