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

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

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

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

(x = 0, x = - 7, y = 0 y = 0)

Solution steps

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

Expand

2 (x - y i) + 5 (x + y i) + (x + y i)^{2} : (x^{2} + 7 x - y^{2}) + i (3 y + 2 x y)

(x^{2} + 7 x - y^{2}) + i (3 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} + 7 x - y^{2}) + i (3 y + 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} + 7 x - y^{2} = 0 3 y + 2 x y = 0]

[x^{2} + 7 x - y^{2} = 0 3 y + 2 x y = 0] : (x = 0, x = - 7, y = 0 y = 0)

(x = 0, x = - 7, y = 0 y = 0)

- 0.008 i = \frac{1}{x \cdot i}

M = C (1 + i)

5 (a + bi) = (a - bi) (- 3 + 4 i)

(1 + i 3) z^{2} + (5 + i 6) z + 7 - i 6 = 0

a^{2} + b^{2} - (a + bi) = 1

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