Simplify (9-i^7)(3+i^5)

$(9 - i^{7}) (3 + i^{5})$

Step 1

Simplify terms.

Simplify each term.

Rewrite $i^{7}$ as $i^{4} (i^{2} \cdot i)$ .

Factor out $i^{4}$ .

$(9 - (i^{4} i^{3})) (3 + i^{5})$

Factor out $i^{2}$ .

$(9 - (i^{4} (i^{2} \cdot i))) (3 + i^{5})$

$(9 - (i^{4} (i^{2} \cdot i))) (3 + i^{5})$

Rewrite $i^{4}$ as $1$ .

Rewrite $i^{4}$ as ${(i^{2})}^{2}$ .

$(9 - ({(i^{2})}^{2} (i^{2} \cdot i))) (3 + i^{5})$

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

$(9 - ({(- 1)}^{2} (i^{2} \cdot i))) (3 + i^{5})$

Raise $- 1$ to the power of $2$ .

$(9 - (1 (i^{2} \cdot i))) (3 + i^{5})$

$(9 - (1 (i^{2} \cdot i))) (3 + i^{5})$

Multiply $i^{2} \cdot i$ by $1$ .

$(9 - (i^{2} \cdot i)) (3 + i^{5})$

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

$(9 - (- 1 \cdot i)) (3 + i^{5})$

Rewrite $- 1 i$ as $- i$ .

$(9 - - i) (3 + i^{5})$

Multiply $- 1$ by $- 1$ .

$(9 + 1 i) (3 + i^{5})$

Multiply $i$ by $1$ .

$(9 + i) (3 + i^{5})$

$(9 + i) (3 + i^{5})$

Simplify each term.

Factor out $i^{4}$ .

$(9 + i) (3 + i^{4} i)$

Rewrite $i^{4}$ as $1$ .

Rewrite $i^{4}$ as ${(i^{2})}^{2}$ .

$(9 + i) (3 + {(i^{2})}^{2} i)$

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

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

Raise $- 1$ to the power of $2$ .

$(9 + i) (3 + 1 i)$

$(9 + i) (3 + 1 i)$

Multiply $i$ by $1$ .

$(9 + i) (3 + i)$

$(9 + i) (3 + i)$

$(9 + i) (3 + i)$

Step 2

Expand $(9 + i) (3 + i)$ using the FOIL Method.

Apply the distributive property.

$9 (3 + i) + i (3 + i)$

Apply the distributive property.

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

Apply the distributive property.

$9 \cdot 3 + 9 i + i \cdot 3 + i i$

$9 \cdot 3 + 9 i + i \cdot 3 + i i$

Step 3

Simplify and combine like terms.

Simplify each term.

Multiply $9$ by $3$ .

$27 + 9 i + i \cdot 3 + i i$

Move $3$ to the left of $i$ .

$27 + 9 i + 3 \cdot i + i i$

Multiply $i i$ .

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

$27 + 9 i + 3 i + i^{1} i$

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

$27 + 9 i + 3 i + i^{1} i^{1}$

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

$27 + 9 i + 3 i + i^{1 + 1}$

Add $1$ and $1$ .

$27 + 9 i + 3 i + i^{2}$

$27 + 9 i + 3 i + i^{2}$

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

$27 + 9 i + 3 i - 1$

$27 + 9 i + 3 i - 1$

Subtract $1$ from $27$ .

$26 + 9 i + 3 i$

Add $9 i$ and $3 i$ .

$26 + 12 i$

$26 + 12 i$

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