What is the answer to (x+i)^2=1 ?

The answer to (x+i)^2=1 is x=1-i,x=-1-i

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(x+i)^2=1

(x + i)^{2} = 1

x = 1 - i, x = - 1 - i

Solution steps

(x + i)^{2} = 1

Substitute

x = a + bi

(a + bi + i)^{2} = 1

Expand

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

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

Rewrite

1

in standard complex form:

1 + 0 i

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

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

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

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

Substitute back

x = a + bi

x = 1 - i, x = - 1 - i

(x + i)^{2} = 2

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

z^{2} - z + 1 = i

i = \frac{57}{98 + j 50.894}

(x + i)^{2} = y