What is the answer to z^2+2z+1+i=0 ?

The answer to z^2+2z+1+i=0 is z=-(-sqrt(2)+2)/2-1/(sqrt(2))i,z=-(sqrt(2)+2)/2+1/(sqrt(2))i

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z^2+2z+1+i=0

z^{2} + 2 z + 1 + i = 0

z = - \frac{- 2 + 2}{2} - \frac{1}{2} i, z = - \frac{2 + 2}{2} + \frac{1}{2} i

Solution steps

z^{2} + 2 z + 1 + i = 0

Substitute

z = x + y i

(x + y i)^{2} + 2 (x + y i) + 1 + i = 0

Expand

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

(x^{2} - y^{2} + 2 x + 1) + i (1 + 2 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} + 2 x + 1) + i (1 + 2 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} - y^{2} + 2 x + 1 = 0 1 + 2 y + 2 x y = 0]

[x^{2} - y^{2} + 2 x + 1 = 0 1 + 2 y + 2 x y = 0] : (x = - \frac{- 2 + 2}{2}, x = - \frac{2 + 2}{2}, y = - \frac{1}{2} y = \frac{1}{2})

Substitute back

z = x + y i

z = - \frac{- 2 + 2}{2} - \frac{1}{2} i, z = - \frac{2 + 2}{2} + \frac{1}{2} i

z^{- 1} = \frac{1}{3 + 2 i}

z^{2} - 2 i z - 3 = 0

x + i y = 0

(x + 2) + (2 y - 1) i = - 1 + 5 i

z^{2} = 1 + i

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