What is the answer to solvefor m,(m-n)^2=-2i ?

The answer to solvefor m,(m-n)^2=-2i is m=(-1+n)+i,m=(n+1)-i

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solvefor m,(m-n)^2=-2i

solve for

m, (m - n)^{2} = - 2 i

m = (- 1 + n) + i, m = (n + 1) - i

Solution steps

(m - n)^{2} = - 2 i

Substitute

m = x + y i

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

Expand

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

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

Rewrite

- 2 i

in standard complex form:

0 - 2 i

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

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

[- 2 n x + x^{2} + n^{2} - y^{2} = 0 - 2 n y + 2 x y = - 2]

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

Substitute back

m = x + y i

m = (- 1 + n) + i, m = (n + 1) - i

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

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

What is the answer to solvefor m,(m-n)^2=-2i ?
The answer to solvefor m,(m-n)^2=-2i is m=(-1+n)+i,m=(n+1)-i