What is the answer to solvefor x,(ix-iy)^2=2 ?

The answer to solvefor x,(ix-iy)^2=2 is x=y+sqrt(2)i,x=y-sqrt(2)i

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solvefor x,(ix-iy)^2=2

solve for

x, (i x - i y)^{2} = 2

x = y + 2 i, x = y - 2 i

Solution steps

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

Substitute

x = a + bi

(i (a + bi) - i y)^{2} = 2

Expand

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

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

Rewrite

2

in standard complex form:

2 + 0 i

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

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

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

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

Substitute back

x = a + bi

x = y + 2 i, x = y - 2 i

(x + i y)^{2} + 2 (x - i y) = - 1 + 6 i

z^{2} = - 2 - 23 i

\frac{i}{z} = - 6

z - \frac{1}{z} = 4 + 3 i

∣ i + z ∣^{2} - i - 2 = z

What is the answer to solvefor x,(ix-iy)^2=2 ?
The answer to solvefor x,(ix-iy)^2=2 is x=y+sqrt(2)i,x=y-sqrt(2)i