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

The answer to z^2+(i-z)z+(3-i)=0 is z=1+3i

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z^2+(i-z)z+(3-i)=0

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

z = 1 + 3 i

Solution steps

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

Substitute

z = x + y i

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

Expand

(x + y i)^{2} + (i - (x + y i)) (x + y i) + 3 - i : (- y + 3) + i (- 1 + x)

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

[- y + 3 = 0 - 1 + x = 0]

[- y + 3 = 0 - 1 + x = 0] : x = 1, y = 3

Substitute back

z = x + y i

z = 1 + 3 i

k_{1} - 3 - 27 i k_{2} = 0

c, i = \frac{d}{( 1 - c )}

(a + i)^{2} = 8 + bi

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

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

What is the answer to z^2+(i-z)z+(3-i)=0 ?
The answer to z^2+(i-z)z+(3-i)=0 is z=1+3i