What is the answer to (x+yi)^2-3i(x+yi)+5i=0 ?

The answer to (x+yi)^2-3i(x+yi)+5i=0 is

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(x+yi)^2-3i(x+yi)+5i=0

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

(x = 1.27140 \dots, x = - 1.27140 \dots, y = - 0.46633 \dots y = 3.46633 \dots)

Solution steps

(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 = 1.27140 \dots, x = - 1.27140 \dots, y = - 0.46633 \dots y = 3.46633 \dots)

(x = 1.27140 \dots, x = - 1.27140 \dots, y = - 0.46633 \dots y = 3.46633 \dots)

x^{2} + 8 i = 0

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

w^{2} - i (w - 2) = (w - 2)

x^{2} - y^{2} + i 2 x y = y + i x

\frac{- 4 + x i}{2 - 3 i} = y - 2 i

What is the answer to (x+yi)^2-3i(x+yi)+5i=0 ?
The answer to (x+yi)^2-3i(x+yi)+5i=0 is