Simplify (1+3i)^2*(2+i)

${(1 + 3 i)}^{2} \cdot (2 + i)$

Step 1

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

$(1 + 3 i) (1 + 3 i) \cdot (2 + i)$

Step 2

Expand $(1 + 3 i) (1 + 3 i)$ using the FOIL Method.

Apply the distributive property.

$(1 (1 + 3 i) + 3 i (1 + 3 i)) \cdot (2 + i)$

Apply the distributive property.

$(1 \cdot 1 + 1 (3 i) + 3 i (1 + 3 i)) \cdot (2 + i)$

Apply the distributive property.

$(1 \cdot 1 + 1 (3 i) + 3 i \cdot 1 + 3 i (3 i)) \cdot (2 + i)$

Step 3

Simplify and combine like terms.

Simplify each term.

Multiply $1$ by $1$ .

$(1 + 1 (3 i) + 3 i \cdot 1 + 3 i (3 i)) \cdot (2 + i)$

Multiply $3 i$ by $1$ .

$(1 + 3 i + 3 i \cdot 1 + 3 i (3 i)) \cdot (2 + i)$

Multiply $3$ by $1$ .

$(1 + 3 i + 3 i + 3 i (3 i)) \cdot (2 + i)$

Multiply $3 i (3 i)$ .

Multiply $3$ by $3$ .

$(1 + 3 i + 3 i + 9 i i) \cdot (2 + i)$

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

$(1 + 3 i + 3 i + 9 (i^{1} i)) \cdot (2 + i)$

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

$(1 + 3 i + 3 i + 9 (i^{1} i^{1})) \cdot (2 + i)$

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

$(1 + 3 i + 3 i + 9 i^{1 + 1}) \cdot (2 + i)$

Add $1$ and $1$ .

$(1 + 3 i + 3 i + 9 i^{2}) \cdot (2 + i)$

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

$(1 + 3 i + 3 i + 9 \cdot - 1) \cdot (2 + i)$

Multiply $9$ by $- 1$ .

$(1 + 3 i + 3 i - 9) \cdot (2 + i)$

Subtract $9$ from $1$ .

$(- 8 + 3 i + 3 i) \cdot (2 + i)$

Add $3 i$ and $3 i$ .

$(- 8 + 6 i) \cdot (2 + i)$

Step 4

Expand $(- 8 + 6 i) (2 + i)$ using the FOIL Method.

Apply the distributive property.

$- 8 (2 + i) + 6 i (2 + i)$

Apply the distributive property.

$- 8 \cdot 2 - 8 i + 6 i (2 + i)$

Apply the distributive property.

$- 8 \cdot 2 - 8 i + 6 i \cdot 2 + 6 i i$

Step 5

Simplify and combine like terms.

Simplify each term.

Multiply $- 8$ by $2$ .

$- 16 - 8 i + 6 i \cdot 2 + 6 i i$

Multiply $2$ by $6$ .

$- 16 - 8 i + 12 i + 6 i i$

Multiply $6 i i$ .

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

$- 16 - 8 i + 12 i + 6 (i^{1} i)$

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

$- 16 - 8 i + 12 i + 6 (i^{1} i^{1})$

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

$- 16 - 8 i + 12 i + 6 i^{1 + 1}$

Add $1$ and $1$ .

$- 16 - 8 i + 12 i + 6 i^{2}$

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

$- 16 - 8 i + 12 i + 6 \cdot - 1$

Multiply $6$ by $- 1$ .

$- 16 - 8 i + 12 i - 6$

Subtract $6$ from $- 16$ .

$- 22 - 8 i + 12 i$

Add $- 8 i$ and $12 i$ .

$- 22 + 4 i$

Simplify (1+3i)^2*(2+i)

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