Write in Standard Form (2+3i)/(1-i)

$\frac{2 + 3 i}{1 - i}$

Step 1

Multiply the numerator and denominator of $\frac{2 + 3 i}{1 - i}$ by the conjugate of $1 - i$ to make the denominator real.

$\frac{2 + 3 i}{1 - i} \cdot \frac{1 + i}{1 + i}$

Step 2

Multiply.

Combine.

$\frac{(2 + 3 i) (1 + i)}{(1 - i) (1 + i)}$

Simplify the numerator.

Expand $(2 + 3 i) (1 + i)$ using the FOIL Method.

Apply the distributive property.

$\frac{2 (1 + i) + 3 i (1 + i)}{(1 - i) (1 + i)}$

Apply the distributive property.

$\frac{2 \cdot 1 + 2 i + 3 i (1 + i)}{(1 - i) (1 + i)}$

Apply the distributive property.

$\frac{2 \cdot 1 + 2 i + 3 i \cdot 1 + 3 i i}{(1 - i) (1 + i)}$

Simplify and combine like terms.

Simplify each term.

Multiply $2$ by $1$ .

$\frac{2 + 2 i + 3 i \cdot 1 + 3 i i}{(1 - i) (1 + i)}$

Multiply $3$ by $1$ .

$\frac{2 + 2 i + 3 i + 3 i i}{(1 - i) (1 + i)}$

Multiply $3 i i$ .

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

$\frac{2 + 2 i + 3 i + 3 \cdot - 1}{(1 - i) (1 + i)}$

Multiply $3$ by $- 1$ .

$\frac{2 + 2 i + 3 i - 3}{(1 - i) (1 + i)}$

Subtract $3$ from $2$ .

$\frac{- 1 + 2 i + 3 i}{(1 - i) (1 + i)}$

Add $2 i$ and $3 i$ .

$\frac{- 1 + 5 i}{(1 - i) (1 + i)}$

Simplify the denominator.

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

Apply the distributive property.

$\frac{- 1 + 5 i}{1 (1 + i) - i (1 + i)}$

Apply the distributive property.

$\frac{- 1 + 5 i}{1 \cdot 1 + 1 i - i (1 + i)}$

Apply the distributive property.

$\frac{- 1 + 5 i}{1 \cdot 1 + 1 i - i \cdot 1 - i i}$

Simplify.

Multiply $1$ by $1$ .

$\frac{- 1 + 5 i}{1 + 1 i - i \cdot 1 - i i}$

Multiply $- 1$ by $1$ .

$\frac{- 1 + 5 i}{1 + 1 i - i - i i}$

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

$\frac{- 1 + 5 i}{1 + 1 i - i - (i^{1} i)}$

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

$\frac{- 1 + 5 i}{1 + 1 i - i - (i^{1} i^{1})}$

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

$\frac{- 1 + 5 i}{1 + 1 i - i - i^{1 + 1}}$

Add $1$ and $1$ .

$\frac{- 1 + 5 i}{1 + 1 i - i - i^{2}}$

Subtract $i$ from $1 i$ .

$\frac{- 1 + 5 i}{1 + 0 - i^{2}}$

Add $1$ and $0$ .

$\frac{- 1 + 5 i}{1 - i^{2}}$

Simplify each term.

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

$\frac{- 1 + 5 i}{1 - - 1}$

Multiply $- 1$ by $- 1$ .

$\frac{- 1 + 5 i}{1 + 1}$

Add $1$ and $1$ .

$\frac{- 1 + 5 i}{2}$

Step 3

Split the fraction $\frac{- 1 + 5 i}{2}$ into two fractions.

$\frac{- 1}{2} + \frac{5 i}{2}$

Step 4

Move the negative in front of the fraction.

$- \frac{1}{2} + \frac{5 i}{2}$

Write in Standard Form (2+3i)/(1-i)

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