Write in Standard Form (4+3i)/(3-i)

$\frac{4 + 3 i}{3 - i}$

Step 1

Multiply the numerator and denominator of $\frac{4 + 3 i}{3 - i}$ by the conjugate of $3 - i$ to make the denominator real.

$\frac{4 + 3 i}{3 - i} \cdot \frac{3 + i}{3 + i}$

Step 2

Multiply.

Combine.

$\frac{(4 + 3 i) (3 + i)}{(3 - i) (3 + i)}$

Simplify the numerator.

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

Apply the distributive property.

$\frac{4 (3 + i) + 3 i (3 + i)}{(3 - i) (3 + i)}$

Apply the distributive property.

$\frac{4 \cdot 3 + 4 i + 3 i (3 + i)}{(3 - i) (3 + i)}$

Apply the distributive property.

$\frac{4 \cdot 3 + 4 i + 3 i \cdot 3 + 3 i i}{(3 - i) (3 + i)}$

Simplify and combine like terms.

Simplify each term.

Multiply $4$ by $3$ .

$\frac{12 + 4 i + 3 i \cdot 3 + 3 i i}{(3 - i) (3 + i)}$

Multiply $3$ by $3$ .

$\frac{12 + 4 i + 9 i + 3 i i}{(3 - i) (3 + i)}$

Multiply $3 i i$ .

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

$\frac{12 + 4 i + 9 i + 3 (i^{1} i)}{(3 - i) (3 + i)}$

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

$\frac{12 + 4 i + 9 i + 3 (i^{1} i^{1})}{(3 - i) (3 + i)}$

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

$\frac{12 + 4 i + 9 i + 3 i^{1 + 1}}{(3 - i) (3 + i)}$

Add $1$ and $1$ .

$\frac{12 + 4 i + 9 i + 3 i^{2}}{(3 - i) (3 + i)}$

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

$\frac{12 + 4 i + 9 i + 3 \cdot - 1}{(3 - i) (3 + i)}$

Multiply $3$ by $- 1$ .

$\frac{12 + 4 i + 9 i - 3}{(3 - i) (3 + i)}$

Subtract $3$ from $12$ .

$\frac{9 + 4 i + 9 i}{(3 - i) (3 + i)}$

Add $4 i$ and $9 i$ .

$\frac{9 + 13 i}{(3 - i) (3 + i)}$

Simplify the denominator.

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

Apply the distributive property.

$\frac{9 + 13 i}{3 (3 + i) - i (3 + i)}$

Apply the distributive property.

$\frac{9 + 13 i}{3 \cdot 3 + 3 i - i (3 + i)}$

Apply the distributive property.

$\frac{9 + 13 i}{3 \cdot 3 + 3 i - i \cdot 3 - i i}$

Simplify.

Multiply $3$ by $3$ .

$\frac{9 + 13 i}{9 + 3 i - i \cdot 3 - i i}$

Multiply $3$ by $- 1$ .

$\frac{9 + 13 i}{9 + 3 i - 3 i - i i}$

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

$\frac{9 + 13 i}{9 + 3 i - 3 i - (i^{1} i)}$

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

$\frac{9 + 13 i}{9 + 3 i - 3 i - (i^{1} i^{1})}$

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

$\frac{9 + 13 i}{9 + 3 i - 3 i - i^{1 + 1}}$

Add $1$ and $1$ .

$\frac{9 + 13 i}{9 + 3 i - 3 i - i^{2}}$

Subtract $3 i$ from $3 i$ .

$\frac{9 + 13 i}{9 + 0 - i^{2}}$

Add $9$ and $0$ .

$\frac{9 + 13 i}{9 - i^{2}}$

Simplify each term.

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

$\frac{9 + 13 i}{9 - - 1}$

Multiply $- 1$ by $- 1$ .

$\frac{9 + 13 i}{9 + 1}$

Add $9$ and $1$ .

$\frac{9 + 13 i}{10}$

Step 3

Split the fraction $\frac{9 + 13 i}{10}$ into two fractions.

$\frac{9}{10} + \frac{13 i}{10}$

Write in Standard Form (4+3i)/(3-i)

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