Examples

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

Step 1

Multiply the numerator and denominator of $\frac{- 3}{- 4 + 1 i}$ by the conjugate of $- 4 + 1 i$ to make the denominator real.

$\frac{- 3}{- 4 + 1 i} \cdot \frac{- 4 - i}{- 4 - i}$

Step 2

Multiply.

Tap for more steps...

Step 2.1

Combine.

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

Step 2.2

Simplify the numerator.

Tap for more steps...

Step 2.2.1

Apply the distributive property.

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

Step 2.2.2

Multiply $- 3$ by $- 4$ .

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

Step 2.2.3

Multiply $- 1$ by $- 3$ .

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

Step 2.3

Simplify the denominator.

Tap for more steps...

Step 2.3.1

Expand $(- 4 + 1 i) (- 4 - i)$ using the FOIL Method.

Tap for more steps...

Step 2.3.1.1

Apply the distributive property.

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

Step 2.3.1.2

Apply the distributive property.

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

Step 2.3.1.3

Apply the distributive property.

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

Step 2.3.2

Simplify.

Tap for more steps...

Step 2.3.2.1

Multiply $- 4$ by $- 4$ .

$\frac{12 + 3 i}{16 - 4 (- i) + 1 i \cdot - 4 + 1 i (- i)}$

Step 2.3.2.2

Multiply $- 1$ by $- 4$ .

$\frac{12 + 3 i}{16 + 4 i + 1 i \cdot - 4 + 1 i (- i)}$

Step 2.3.2.3

Multiply $- 4$ by $1$ .

$\frac{12 + 3 i}{16 + 4 i - 4 i + 1 i (- i)}$

Step 2.3.2.4

Multiply $- 1$ by $1$ .

$\frac{12 + 3 i}{16 + 4 i - 4 i - i i}$

Step 2.3.2.5

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

$\frac{12 + 3 i}{16 + 4 i - 4 i - (i^{1} i)}$

Step 2.3.2.6

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

$\frac{12 + 3 i}{16 + 4 i - 4 i - (i^{1} i^{1})}$

Step 2.3.2.7

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

$\frac{12 + 3 i}{16 + 4 i - 4 i - i^{1 + 1}}$

Step 2.3.2.8

Add $1$ and $1$ .

$\frac{12 + 3 i}{16 + 4 i - 4 i - i^{2}}$

Step 2.3.2.9

Subtract $4 i$ from $4 i$ .

$\frac{12 + 3 i}{16 + 0 - i^{2}}$

Step 2.3.2.10

Add $16$ and $0$ .

$\frac{12 + 3 i}{16 - i^{2}}$

Step 2.3.3

Simplify each term.

Tap for more steps...

Step 2.3.3.1

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

$\frac{12 + 3 i}{16 - - 1}$

Step 2.3.3.2

Multiply $- 1$ by $- 1$ .

$\frac{12 + 3 i}{16 + 1}$

Step 2.3.4

Add $16$ and $1$ .

$\frac{12 + 3 i}{17}$

Step 3

Split the fraction $\frac{12 + 3 i}{17}$ into two fractions.

$\frac{12}{17} + \frac{3 i}{17}$

Enter YOUR Problem