Write in Standard Form (3+5i)/(1+i)

$\frac{3 + 5 i}{1 + i}$

Step 1

Multiply the numerator and denominator of $\frac{3 + 5 i}{1 + 1 i}$ by the conjugate of $1 + 1 i$ to make the denominator real.

$\frac{3 + 5 i}{1 + 1 i} \cdot \frac{1 - i}{1 - i}$

Step 2

Multiply.

Combine.

$\frac{(3 + 5 i) (1 - i)}{(1 + 1 i) (1 - i)}$

Simplify the numerator.

Expand $(3 + 5 i) (1 - i)$ using the FOIL Method.

Apply the distributive property.

$\frac{3 (1 - i) + 5 i (1 - i)}{(1 + 1 i) (1 - i)}$

Apply the distributive property.

$\frac{3 \cdot 1 + 3 (- i) + 5 i (1 - i)}{(1 + 1 i) (1 - i)}$

Apply the distributive property.

$\frac{3 \cdot 1 + 3 (- i) + 5 i \cdot 1 + 5 i (- i)}{(1 + 1 i) (1 - i)}$

Simplify and combine like terms.

Simplify each term.

Multiply $3$ by $1$ .

$\frac{3 + 3 (- i) + 5 i \cdot 1 + 5 i (- i)}{(1 + 1 i) (1 - i)}$

Multiply $- 1$ by $3$ .

$\frac{3 - 3 i + 5 i \cdot 1 + 5 i (- i)}{(1 + 1 i) (1 - i)}$

Multiply $5$ by $1$ .

$\frac{3 - 3 i + 5 i + 5 i (- i)}{(1 + 1 i) (1 - i)}$

Multiply $5 i (- i)$ .

Multiply $- 1$ by $5$ .

$\frac{3 - 3 i + 5 i - 5 i i}{(1 + 1 i) (1 - i)}$

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

$\frac{3 - 3 i + 5 i - 5 (i^{1} i)}{(1 + 1 i) (1 - i)}$

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

$\frac{3 - 3 i + 5 i - 5 (i^{1} i^{1})}{(1 + 1 i) (1 - i)}$

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

$\frac{3 - 3 i + 5 i - 5 i^{1 + 1}}{(1 + 1 i) (1 - i)}$

Add $1$ and $1$ .

$\frac{3 - 3 i + 5 i - 5 i^{2}}{(1 + 1 i) (1 - i)}$

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

$\frac{3 - 3 i + 5 i - 5 \cdot - 1}{(1 + 1 i) (1 - i)}$

Multiply $- 5$ by $- 1$ .

$\frac{3 - 3 i + 5 i + 5}{(1 + 1 i) (1 - i)}$

Add $3$ and $5$ .

$\frac{8 - 3 i + 5 i}{(1 + 1 i) (1 - i)}$

Add $- 3 i$ and $5 i$ .

$\frac{8 + 2 i}{(1 + 1 i) (1 - i)}$

Simplify the denominator.

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

Apply the distributive property.

$\frac{8 + 2 i}{1 (1 - i) + 1 i (1 - i)}$

Apply the distributive property.

$\frac{8 + 2 i}{1 \cdot 1 + 1 (- i) + 1 i (1 - i)}$

Apply the distributive property.

$\frac{8 + 2 i}{1 \cdot 1 + 1 (- i) + 1 i \cdot 1 + 1 i (- i)}$

Simplify.

Multiply $1$ by $1$ .

$\frac{8 + 2 i}{1 + 1 (- i) + 1 i \cdot 1 + 1 i (- i)}$

Multiply $- 1$ by $1$ .

$\frac{8 + 2 i}{1 - 1 i + 1 i \cdot 1 + 1 i (- i)}$

Multiply $1$ by $1$ .

$\frac{8 + 2 i}{1 - 1 i + 1 i + 1 i (- i)}$

Multiply $- 1$ by $1$ .

$\frac{8 + 2 i}{1 - 1 i + 1 i - i i}$

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

$\frac{8 + 2 i}{1 - 1 i + 1 i - (i^{1} i)}$

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

$\frac{8 + 2 i}{1 - 1 i + 1 i - (i^{1} i^{1})}$

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

$\frac{8 + 2 i}{1 - 1 i + 1 i - i^{1 + 1}}$

Add $1$ and $1$ .

$\frac{8 + 2 i}{1 - 1 i + 1 i - i^{2}}$

Add $- 1 i$ and $1 i$ .

$\frac{8 + 2 i}{1 + 0 - i^{2}}$

Add $1$ and $0$ .

$\frac{8 + 2 i}{1 - i^{2}}$

Simplify each term.

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

$\frac{8 + 2 i}{1 - - 1}$

Multiply $- 1$ by $- 1$ .

$\frac{8 + 2 i}{1 + 1}$

Add $1$ and $1$ .

$\frac{8 + 2 i}{2}$

Step 3

Cancel the common factor of $8 + 2 i$ and $2$ .

Factor $2$ out of $8$ .

$\frac{2 \cdot 4 + 2 i}{2}$

Factor $2$ out of $2 i$ .

$\frac{2 \cdot 4 + 2 (i)}{2}$

Factor $2$ out of $2 (4) + 2 (i)$ .

$\frac{2 (4 + i)}{2}$

Cancel the common factors.

Factor $2$ out of $2$ .

$\frac{2 (4 + i)}{2 (1)}$

Cancel the common factor.

$\frac{2 (4 + i)}{2 \cdot 1}$

Rewrite the expression.

$\frac{4 + i}{1}$

Divide $4 + i$ by $1$ .

$4 + i$

Write in Standard Form (3+5i)/(1+i)

Related Questions

Support LCMGCF.com by disabling your adblocker.