What is (x^2+2xyi-y^2-2)^2 ?

The solution to (x^2+2xyi-y^2-2)^2 is (x^4+y^4-5x^2y^2-4x^2-y^2x^2+4y^2+4)+(4x^3y-4xy^3-8xy)i

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simplify (x^2+2xyi-y^2-2)^2

simplify

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

(x^{4} + y^{4} - 5 x^{2} y^{2} - 4 x^{2} - y^{2} x^{2} + 4 y^{2} + 4) + (4 x^{3} y - 4 x y^{3} - 8 x y) i

Solution steps

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

Apply exponent rule:

a^{b + c} = a^{b} \cdot a^{c}

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

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

Distribute parentheses

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

Simplify

x^{2} x^{2} + x^{2} \cdot 2 x y i + x^{2} (- y^{2}) + x^{2} (- 2) + 2 x y i x^{2} + 2 x y i 2 x y i + 2 x y i (- y^{2}) + 2 x y i (- 2) - y^{2} x^{2} - y^{2} \cdot 2 x y i - y^{2} (- y^{2}) - y^{2} (- 2) - 2 x^{2} - 2 \cdot 2 x y i - 2 (- y^{2}) - 2 (- 2) : x^{4} + y^{4} - 5 x^{2} y^{2} + 4 x^{3} y i - 4 x^{2} - y^{2} x^{2} - 4 x i y^{3} - 8 x y i + 4 y^{2} + 4

= x^{4} + y^{4} - 5 x^{2} y^{2} + 4 x^{3} y i - 4 x^{2} - y^{2} x^{2} - 4 x i y^{3} - 8 x y i + 4 y^{2} + 4

Rewrite in standard complex form:

(x^{4} + y^{4} - 5 x^{2} y^{2} - 4 x^{2} - y^{2} x^{2} + 4 y^{2} + 4) + (4 x^{3} y - 4 x y^{3} - 8 x y) i

= (x^{4} + y^{4} - 5 x^{2} y^{2} - 4 x^{2} - y^{2} x^{2} + 4 y^{2} + 4) + (4 x^{3} y - 4 x y^{3} - 8 x y) i

\frac{100 \cdot b \cdot z ^{2}}{( 10 \cdot z - 9 ) ^{2}}

(1 + 1 - 4 a)^{2}

- \frac{1}{x ( x + 4 ) ( x + 5 )}

\frac{( x - 2 ) ( x - 3 )}{x + 2}

\frac{1}{k ( k + 1 )} + (k + 1)

What is (x^2+2xyi-y^2-2)^2 ?
The solution to (x^2+2xyi-y^2-2)^2 is (x^4+y^4-5x^2y^2-4x^2-y^2x^2+4y^2+4)+(4x^3y-4xy^3-8xy)i