Simplify (4i)/(2-8i)

$\frac{4 i}{2 - 8 i}$

Cancel the common factor of $4$ and $2 - 8 i$ .

Factor $2$ out of $4 i$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Factor $2$ out of $- 8 i$ .

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

Factor $2$ out of $2 (1) + 2 (- 4 i)$ .

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

Cancel the common factor.

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

Rewrite the expression.

$\frac{2 i}{1 - 4 i}$

Multiply the numerator and denominator of $\frac{2 i}{1 - 4 i}$ by the conjugate of $1 - 4 i$ to make the denominator real.

$\frac{2 i}{1 - 4 i} \cdot \frac{1 + 4 i}{1 + 4 i}$

Multiply.

Combine.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $2$ by $1$ .

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

Multiply $2 i (4 i)$ .

Multiply $4$ by $2$ .

$\frac{2 i + 8 i i}{(1 - 4 i) (1 + 4 i)}$

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

$\frac{2 i + 8 (i^{1} i)}{(1 - 4 i) (1 + 4 i)}$

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

$\frac{2 i + 8 (i^{1} i^{1})}{(1 - 4 i) (1 + 4 i)}$

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

$\frac{2 i + 8 i^{1 + 1}}{(1 - 4 i) (1 + 4 i)}$

Add $1$ and $1$ .

$\frac{2 i + 8 i^{2}}{(1 - 4 i) (1 + 4 i)}$

Simplify each term.

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

$\frac{2 i + 8 \cdot - 1}{(1 - 4 i) (1 + 4 i)}$

Multiply $8$ by $- 1$ .

$\frac{2 i - 8}{(1 - 4 i) (1 + 4 i)}$

Reorder $2 i$ and $- 8$ .

$\frac{- 8 + 2 i}{(1 - 4 i) (1 + 4 i)}$

Simplify the denominator.

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

Apply the distributive property.

$\frac{- 8 + 2 i}{1 (1 + 4 i) - 4 i (1 + 4 i)}$

Apply the distributive property.

$\frac{- 8 + 2 i}{1 \cdot 1 + 1 (4 i) - 4 i (1 + 4 i)}$

Apply the distributive property.

$\frac{- 8 + 2 i}{1 \cdot 1 + 1 (4 i) - 4 i \cdot 1 - 4 i (4 i)}$

Simplify.

Multiply $1$ by $1$ .

$\frac{- 8 + 2 i}{1 + 1 (4 i) - 4 i \cdot 1 - 4 i (4 i)}$

Multiply $4$ by $1$ .

$\frac{- 8 + 2 i}{1 + 4 i - 4 i \cdot 1 - 4 i (4 i)}$

Multiply $- 4$ by $1$ .

$\frac{- 8 + 2 i}{1 + 4 i - 4 i - 4 i (4 i)}$

Multiply $4$ by $- 4$ .

$\frac{- 8 + 2 i}{1 + 4 i - 4 i - 16 i i}$

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

$\frac{- 8 + 2 i}{1 + 4 i - 4 i - 16 (i^{1} i)}$

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

$\frac{- 8 + 2 i}{1 + 4 i - 4 i - 16 (i^{1} i^{1})}$

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

$\frac{- 8 + 2 i}{1 + 4 i - 4 i - 16 i^{1 + 1}}$

Add $1$ and $1$ .

$\frac{- 8 + 2 i}{1 + 4 i - 4 i - 16 i^{2}}$

Subtract $4 i$ from $4 i$ .

$\frac{- 8 + 2 i}{1 + 0 - 16 i^{2}}$

Add $1$ and $0$ .

$\frac{- 8 + 2 i}{1 - 16 i^{2}}$

Simplify each term.

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

$\frac{- 8 + 2 i}{1 - 16 \cdot - 1}$

Multiply $- 16$ by $- 1$ .

$\frac{- 8 + 2 i}{1 + 16}$

Add $1$ and $16$ .

$\frac{- 8 + 2 i}{17}$

Split the fraction $\frac{- 8 + 2 i}{17}$ into two fractions.

$\frac{- 8}{17} + \frac{2 i}{17}$

Move the negative in front of the fraction.

$- \frac{8}{17} + \frac{2 i}{17}$

Simplify (4i)/(2-8i)

Related Questions

Support LCMGCF.com by disabling your adblocker.