Finite Math Examples

${(4 + 3 i)}^{2}$

Step 1

Set ${(4 + 3 i)}^{2}$ equal to $0$ .

${(4 + 3 i)}^{2} = 0$

Step 2

Simplify the left side.

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Step 2.1

Simplify ${(4 + 3 i)}^{2}$ .

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Step 2.1.1

Rewrite ${(4 + 3 i)}^{2}$ as $(4 + 3 i) (4 + 3 i)$ .

$(4 + 3 i) (4 + 3 i) = 0$

Step 2.1.2

Expand $(4 + 3 i) (4 + 3 i)$ using the FOIL Method.

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Step 2.1.2.1

Apply the distributive property.

$4 (4 + 3 i) + 3 i (4 + 3 i) = 0$

Step 2.1.2.2

Apply the distributive property.

$4 \cdot 4 + 4 (3 i) + 3 i (4 + 3 i) = 0$

Step 2.1.2.3

Apply the distributive property.

$4 \cdot 4 + 4 (3 i) + 3 i \cdot 4 + 3 i (3 i) = 0$

Step 2.1.3

Simplify and combine like terms.

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Step 2.1.3.1

Simplify each term.

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Step 2.1.3.1.1

Multiply $4$ by $4$ .

$16 + 4 (3 i) + 3 i \cdot 4 + 3 i (3 i) = 0$

Step 2.1.3.1.2

Multiply $3$ by $4$ .

$16 + 12 i + 3 i \cdot 4 + 3 i (3 i) = 0$

Step 2.1.3.1.3

Multiply $4$ by $3$ .

$16 + 12 i + 12 i + 3 i (3 i) = 0$

Step 2.1.3.1.4

Multiply $3 i (3 i)$ .

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Step 2.1.3.1.4.1

Multiply $3$ by $3$ .

$16 + 12 i + 12 i + 9 i i = 0$

Step 2.1.3.1.4.2

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

$16 + 12 i + 12 i + 9 (i^{1} i) = 0$

Step 2.1.3.1.4.3

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

$16 + 12 i + 12 i + 9 (i^{1} i^{1}) = 0$

Step 2.1.3.1.4.4

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

$16 + 12 i + 12 i + 9 i^{1 + 1} = 0$

Step 2.1.3.1.4.5

Add $1$ and $1$ .

$16 + 12 i + 12 i + 9 i^{2} = 0$

Step 2.1.3.1.5

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

$16 + 12 i + 12 i + 9 \cdot - 1 = 0$

Step 2.1.3.1.6

Multiply $9$ by $- 1$ .

$16 + 12 i + 12 i - 9 = 0$

Step 2.1.3.2

Subtract $9$ from $16$ .

$7 + 12 i + 12 i = 0$

Step 2.1.3.3

Add $12 i$ and $12 i$ .

$7 + 24 i = 0$

Step 3

Since $7 + 24 i \neq 0$ , there are no solutions.

No solution