Solve for x x+yi=(a+bi)^2

$x + y i = {(a + b i)}^{2}$

Simplify ${(a + b i)}^{2}$ .

Rewrite.

$x + y i = 0 + 0 + {(a + b i)}^{2}$

Rewrite ${(a + b i)}^{2}$ as $(a + b i) (a + b i)$ .

$x + y i = (a + b i) (a + b i)$

Expand $(a + b i) (a + b i)$ using the FOIL Method.

Apply the distributive property.

$x + y i = a (a + b i) + b i (a + b i)$

Apply the distributive property.

$x + y i = a \cdot a + a (b i) + b i (a + b i)$

Apply the distributive property.

$x + y i = a \cdot a + a (b i) + b i a + b i (b i)$

Simplify and combine like terms.

Simplify each term.

Multiply $a$ by $a$ .

$x + y i = a^{2} + a (b i) + b i a + b i (b i)$

Rewrite using the commutative property of multiplication.

$x + y i = a^{2} + i a b + b i a + b i (b i)$

Multiply $b$ by $b$ by adding the exponents.

Move $b$ .

$x + y i = a^{2} + i a b + b i a + b \cdot b i i$

Multiply $b$ by $b$ .

$x + y i = a^{2} + i a b + b i a + b^{2} i i$

Multiply $b^{2} i i$ .

Raise $i$ to the power of $1$ .

$x + y i = a^{2} + i a b + b i a + b^{2} (i^{1} i)$

Raise $i$ to the power of $1$ .

$x + y i = a^{2} + i a b + b i a + b^{2} (i^{1} i^{1})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x + y i = a^{2} + i a b + b i a + b^{2} i^{1 + 1}$

Add $1$ and $1$ .

$x + y i = a^{2} + i a b + b i a + b^{2} i^{2}$

Rewrite $i^{2}$ as $- 1$ .

$x + y i = a^{2} + i a b + b i a + b^{2} \cdot - 1$

Move $- 1$ to the left of $b^{2}$ .

$x + y i = a^{2} + i a b + b i a - 1 \cdot b^{2}$

Rewrite $- 1 b^{2}$ as $- b^{2}$ .

$x + y i = a^{2} + i a b + b i a - b^{2}$

Reorder $b$ and $a$ .

$x + y i = a^{2} + a b i + a b i - b^{2}$

Add $a b i$ and $a b i$ .

$x + y i = a^{2} + 2 a b i - b^{2}$

Subtract $y i$ from both sides of the equation.

$x = a^{2} + 2 a b i - b^{2} - y i$