Evaluate (2-3i)^-1

${(2 - 3 i)}^{- 1}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

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

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

Multiply.

Combine.

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

Multiply $2 + 3 i$ by $1$ .

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

Simplify the denominator.

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

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Simplify.

Multiply $2$ by $2$ .

$\frac{2 + 3 i}{4 + 2 (3 i) - 3 i \cdot 2 - 3 i (3 i)}$

Multiply $3$ by $2$ .

$\frac{2 + 3 i}{4 + 6 i - 3 i \cdot 2 - 3 i (3 i)}$

Multiply $2$ by $- 3$ .

$\frac{2 + 3 i}{4 + 6 i - 6 i - 3 i (3 i)}$

Multiply $3$ by $- 3$ .

$\frac{2 + 3 i}{4 + 6 i - 6 i - 9 i i}$

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

$\frac{2 + 3 i}{4 + 6 i - 6 i - 9 (i^{1} i)}$

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

$\frac{2 + 3 i}{4 + 6 i - 6 i - 9 (i^{1} i^{1})}$

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

$\frac{2 + 3 i}{4 + 6 i - 6 i - 9 i^{1 + 1}}$

Add $1$ and $1$ .

$\frac{2 + 3 i}{4 + 6 i - 6 i - 9 i^{2}}$

Subtract $6 i$ from $6 i$ .

$\frac{2 + 3 i}{4 + 0 - 9 i^{2}}$

Add $4$ and $0$ .

$\frac{2 + 3 i}{4 - 9 i^{2}}$

Simplify each term.

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

$\frac{2 + 3 i}{4 - 9 \cdot - 1}$

Multiply $- 9$ by $- 1$ .

$\frac{2 + 3 i}{4 + 9}$

Add $4$ and $9$ .

$\frac{2 + 3 i}{13}$

Split the fraction $\frac{2 + 3 i}{13}$ into two fractions.

$\frac{2}{13} + \frac{3 i}{13}$

Evaluate (2-3i)^-1

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