Solve for a (1+ square root of 3i)^2=a+ib

${(1 + \sqrt{3} i)}^{2} = a + i b$

Rewrite the equation as $a + i b = {(1 + \sqrt{3} i)}^{2}$ .

$a + i b = {(1 + \sqrt{3} i)}^{2}$

Simplify ${(1 + \sqrt{3} i)}^{2}$ .

Rewrite.

$a + i b = 0 + 0 + {(1 + \sqrt{3} i)}^{2}$

Rewrite ${(1 + \sqrt{3} i)}^{2}$ as $(1 + \sqrt{3} i) (1 + \sqrt{3} i)$ .

$a + i b = (1 + \sqrt{3} i) (1 + \sqrt{3} i)$

Expand $(1 + \sqrt{3} i) (1 + \sqrt{3} i)$ using the FOIL Method.

Apply the distributive property.

$a + i b = 1 (1 + \sqrt{3} i) + \sqrt{3} i (1 + \sqrt{3} i)$

Apply the distributive property.

$a + i b = 1 \cdot 1 + 1 (\sqrt{3} i) + \sqrt{3} i (1 + \sqrt{3} i)$

Apply the distributive property.

$a + i b = 1 \cdot 1 + 1 (\sqrt{3} i) + \sqrt{3} i \cdot 1 + \sqrt{3} i (\sqrt{3} i)$

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$a + i b = 1 + 1 (\sqrt{3} i) + \sqrt{3} i \cdot 1 + \sqrt{3} i (\sqrt{3} i)$

Multiply $\sqrt{3} i$ by $1$ .

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i \cdot 1 + \sqrt{3} i (\sqrt{3} i)$

Multiply $\sqrt{3}$ by $1$ .

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + \sqrt{3} i (\sqrt{3} i)$

Multiply $\sqrt{3} i (\sqrt{3} i)$ .

Raise $\sqrt{3}$ to the power of $1$ .

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + {\sqrt{3}}^{1} \sqrt{3} i i$

Raise $\sqrt{3}$ to the power of $1$ .

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + {\sqrt{3}}^{1} {\sqrt{3}}^{1} i i$

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

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + {\sqrt{3}}^{1 + 1} i i$

Add $1$ and $1$ .

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + {\sqrt{3}}^{2} i i$

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

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + {\sqrt{3}}^{2} (i^{1} i)$

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

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + {\sqrt{3}}^{2} (i^{1} i^{1})$

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

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + {\sqrt{3}}^{2} i^{1 + 1}$

Add $1$ and $1$ .

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + {\sqrt{3}}^{2} i^{2}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + {(3^{\frac{1}{2}})}^{2} i^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + 3^{\frac{1}{2} \cdot 2} i^{2}$

Combine $\frac{1}{2}$ and $2$ .

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + 3^{\frac{2}{2}} i^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + 3^{\frac{2}{2}} i^{2}$

Rewrite the expression.

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + 3^{1} i^{2}$

Evaluate the exponent.

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + 3 i^{2}$

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

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i + 3 \cdot - 1$

Multiply $3$ by $- 1$ .

$a + i b = 1 + \sqrt{3} i + \sqrt{3} i - 3$

Subtract $3$ from $1$ .

$a + i b = - 2 + \sqrt{3} i + \sqrt{3} i$

Add $\sqrt{3} i$ and $\sqrt{3} i$ .

$a + i b = - 2 + 2 \sqrt{3} i$

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

$a = - 2 + 2 \sqrt{3} i - i b$

Solve for a (1+ square root of 3i)^2=a+ib

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