Simplify (1+i)^4

${(1 + i)}^{4}$

Step 1

Use the Binomial Theorem.

$1^{4} + 4 \cdot 1^{3} i + 6 \cdot 1^{2} i^{2} + 4 \cdot 1 i^{3} + i^{4}$

Step 2

Simplify terms.

Simplify each term.

One to any power is one.

$1 + 4 \cdot 1^{3} i + 6 \cdot 1^{2} i^{2} + 4 \cdot 1 i^{3} + i^{4}$

One to any power is one.

$1 + 4 \cdot 1 i + 6 \cdot 1^{2} i^{2} + 4 \cdot 1 i^{3} + i^{4}$

Multiply $4$ by $1$ .

$1 + 4 i + 6 \cdot 1^{2} i^{2} + 4 \cdot 1 i^{3} + i^{4}$

One to any power is one.

$1 + 4 i + 6 \cdot 1 i^{2} + 4 \cdot 1 i^{3} + i^{4}$

Multiply $6$ by $1$ .

$1 + 4 i + 6 i^{2} + 4 \cdot 1 i^{3} + i^{4}$

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

$1 + 4 i + 6 \cdot - 1 + 4 \cdot 1 i^{3} + i^{4}$

Multiply $6$ by $- 1$ .

$1 + 4 i - 6 + 4 \cdot 1 i^{3} + i^{4}$

Multiply $4$ by $1$ .

$1 + 4 i - 6 + 4 i^{3} + i^{4}$

Factor out $i^{2}$ .

$1 + 4 i - 6 + 4 (i^{2} \cdot i) + i^{4}$

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

$1 + 4 i - 6 + 4 (- 1 \cdot i) + i^{4}$

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

$1 + 4 i - 6 + 4 (- i) + i^{4}$

Multiply $- 1$ by $4$ .

$1 + 4 i - 6 - 4 i + i^{4}$

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

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

$1 + 4 i - 6 - 4 i + {(i^{2})}^{2}$

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

$1 + 4 i - 6 - 4 i + {(- 1)}^{2}$

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

$1 + 4 i - 6 - 4 i + 1$

Simplify by adding terms.

Subtract $6$ from $1$ .

$- 5 + 4 i - 4 i + 1$

Add $- 5$ and $1$ .

$- 4 + 4 i - 4 i$

Subtract $4 i$ from $4 i$ .

$- 4 + 0$

Add $- 4$ and $0$ .

$- 4$

Simplify (1+i)^4

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