What is the answer to z^2-3iz+5i=0 ?

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z^2-3iz+5i=0

z^{2} - 3 i z + 5 i = 0

z = \frac{- 9 + 481}{2 2} + \frac{- 10 2 + 3 - 9 + 481}{2 - 9 + 481} i, z = - \frac{- 9 + 481}{2 2} + \frac{3 481 - 9 + 10 2}{2 - 9 + 481} i

Solution steps

z^{2} - 3 i z + 5 i = 0

Substitute

z = x + y i

(x + y i)^{2} - 3 i (x + y i) + 5 i = 0

Expand

(x + y i)^{2} - 3 i (x + y i) + 5 i : (x^{2} - y^{2} + 3 y) + i (5 - 3 x + 2 x y)

(x^{2} - y^{2} + 3 y) + i (5 - 3 x + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} + 3 y) + i (5 - 3 x + 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} + 3 y = 0 5 - 3 x + 2 x y = 0]

[x^{2} - y^{2} + 3 y = 0 5 - 3 x + 2 x y = 0] : x = \frac{- 9 + 481}{2 2}, x = - \frac{- 9 + 481}{2 2}, y = \frac{- 10 2 + 3 - 9 + 481}{2 - 9 + 481} y = \frac{3 481 - 9 + 10 2}{2 - 9 + 481}

Substitute back

z = x + y i

z = \frac{- 9 + 481}{2 2} + \frac{- 10 2 + 3 - 9 + 481}{2 - 9 + 481} i, z = - \frac{- 9 + 481}{2 2} + \frac{3 481 - 9 + 10 2}{2 - 9 + 481} i

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

x, cs in y + x^{2} y = c

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

4 z^{2} + (- 13 + 18 i) z - 5 - 10 i = 0

x^{2} + 4 - 3 i = 0