Simplify ( square root of -4)/((5+2i)-(3-4i))

$\frac{\sqrt{- 4}}{(5 + 2 i) - (3 - 4 i)}$

Step 1

Pull out imaginary unit $i$ .

Rewrite $- 4$ as $- 1 (4)$ .

$\frac{\sqrt{- 1 (4)}}{(5 + 2 i) - (3 - 4 i)}$

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$\frac{\sqrt{- 1} \cdot \sqrt{4}}{(5 + 2 i) - (3 - 4 i)}$

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

$\frac{i \cdot \sqrt{4}}{(5 + 2 i) - (3 - 4 i)}$

Step 2

Simplify the numerator.

Rewrite $4$ as $2^{2}$ .

$\frac{i \sqrt{2^{2}}}{5 + 2 i - (3 - 4 i)}$

Pull terms out from under the radical, assuming positive real numbers.

$\frac{i \cdot 2}{5 + 2 i - (3 - 4 i)}$

Step 3

Simplify the denominator.

Apply the distributive property.

$\frac{i \cdot 2}{5 + 2 i - 1 \cdot 3 - (- 4 i)}$

Multiply $- 1$ by $3$ .

$\frac{i \cdot 2}{5 + 2 i - 3 - (- 4 i)}$

Multiply $- 4$ by $- 1$ .

$\frac{i \cdot 2}{5 + 2 i - 3 + 4 i}$

Subtract $3$ from $5$ .

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

Add $2 i$ and $4 i$ .

$\frac{i \cdot 2}{2 + 6 i}$

Step 4

Reduce the expression by cancelling the common factors.

Move $2$ to the left of $i$ .

$\frac{2 i}{2 + 6 i}$

Cancel the common factor of $2$ and $2 + 6 i$ .

Factor $2$ out of $2 i$ .

$\frac{2 (i)}{2 + 6 i}$

Cancel the common factors.

Factor $2$ out of $2$ .

$\frac{2 (i)}{2 (1) + 6 i}$

Factor $2$ out of $6 i$ .

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

Factor $2$ out of $2 (1) + 2 (3 i)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 5

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

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

Step 6

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 $9$ .

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

Step 7

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

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

Simplify ( square root of -4)/((5+2i)-(3-4i))

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