What is the answer to (a+bi)^2+2(a-bi)+1=0 ?

The answer to (a+bi)^2+2(a-bi)+1=0 is

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(a+bi)^2+2(a-bi)+1=0

(a + bi)^{2} + 2 (a - bi) + 1 = 0

a = - 1, a = 1, a = 1, b = 0 b = 2 b = - 2

Solution steps

(a + bi)^{2} + 2 (a - bi) + 1 = 0

Expand

(a + bi)^{2} + 2 (a - bi) + 1 : (a^{2} - b^{2} + 2 a + 1) + i (- 2 b + 2 ab)

(a^{2} - b^{2} + 2 a + 1) + i (- 2 b + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + 2 a + 1) + i (- 2 b + 2 ab) = 0 + 0 i

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

[a^{2} - b^{2} + 2 a + 1 = 0 - 2 b + 2 ab = 0]

[a^{2} - b^{2} + 2 a + 1 = 0 - 2 b + 2 ab = 0] : a = - 1, a = 1, a = 1, b = 0 b = 2 b = - 2

a = - 1, a = 1, a = 1, b = 0 b = 2 b = - 2

m, y = (x - 3)^{2} minim o

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

2 - i x - y = 0

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

(x + y) - (3 x + 2 y) i = 5 + 2 i

What is the answer to (a+bi)^2+2(a-bi)+1=0 ?
The answer to (a+bi)^2+2(a-bi)+1=0 is