Simplify ( square root of -1)/((3+8i)-(2+5i))

$\frac{\sqrt{- 1}}{(3 + 8 i) - (2 + 5 i)}$

Rewrite $\sqrt{- 1}$ as $i$ .

$\frac{i}{(3 + 8 i) - (2 + 5 i)}$

Simplify the denominator.

Apply the distributive property.

$\frac{i}{3 + 8 i - 1 \cdot 2 - (5 i)}$

Multiply $- 1$ by $2$ .

$\frac{i}{3 + 8 i - 2 - (5 i)}$

Multiply $5$ by $- 1$ .

$\frac{i}{3 + 8 i - 2 - 5 i}$

Subtract $2$ from $3$ .

$\frac{i}{1 + 8 i - 5 i}$

Subtract $5 i$ from $8 i$ .

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

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

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

Multiply.

Combine.

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $i$ by $1$ .

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

Multiply $i (- 3 i)$ .

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

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

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

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

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

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

Add $1$ and $1$ .

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

Simplify each term.

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

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

Multiply $- 3$ by $- 1$ .

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

Reorder $i$ and $3$ .

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

Simplify the denominator.

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify.

Multiply $1$ by $1$ .

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

Multiply $- 3$ by $1$ .

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

Multiply $3$ by $1$ .

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

Multiply $- 3$ by $3$ .

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

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

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

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

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

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

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

Add $1$ and $1$ .

$\frac{3 + i}{1 - 3 i + 3 i - 9 i^{2}}$

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

$\frac{3 + i}{1 + 0 - 9 i^{2}}$

Add $1$ and $0$ .

$\frac{3 + i}{1 - 9 i^{2}}$

Simplify each term.

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

$\frac{3 + i}{1 - 9 \cdot - 1}$

Multiply $- 9$ by $- 1$ .

$\frac{3 + i}{1 + 9}$

Add $1$ and $9$ .

$\frac{3 + i}{10}$

Split the fraction $\frac{3 + i}{10}$ into two fractions.

$\frac{3}{10} + \frac{i}{10}$

Simplify ( square root of -1)/((3+8i)-(2+5i))

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