Solve for a (15-4i)(6-3i)=a+bi

$(15 - 4 i) (6 - 3 i) = a + b i$

Step 1

Rewrite the equation as $a + b i = (15 - 4 i) (6 - 3 i)$ .

$a + b i = (15 - 4 i) (6 - 3 i)$

Step 2

Simplify $(15 - 4 i) (6 - 3 i)$ .

Rewrite.

$a + b i = 0 + 0 + (15 - 4 i) (6 - 3 i)$

Simplify by adding zeros.

$a + b i = (15 - 4 i) (6 - 3 i)$

Expand $(15 - 4 i) (6 - 3 i)$ using the FOIL Method.

Apply the distributive property.

$a + b i = 15 (6 - 3 i) - 4 i (6 - 3 i)$

Apply the distributive property.

$a + b i = 15 \cdot 6 + 15 (- 3 i) - 4 i (6 - 3 i)$

Apply the distributive property.

$a + b i = 15 \cdot 6 + 15 (- 3 i) - 4 i \cdot 6 - 4 i (- 3 i)$

Simplify and combine like terms.

Simplify each term.

Multiply $15$ by $6$ .

$a + b i = 90 + 15 (- 3 i) - 4 i \cdot 6 - 4 i (- 3 i)$

Multiply $- 3$ by $15$ .

$a + b i = 90 - 45 i - 4 i \cdot 6 - 4 i (- 3 i)$

Multiply $6$ by $- 4$ .

$a + b i = 90 - 45 i - 24 i - 4 i (- 3 i)$

Multiply $- 4 i (- 3 i)$ .

Multiply $- 3$ by $- 4$ .

$a + b i = 90 - 45 i - 24 i + 12 i i$

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

$a + b i = 90 - 45 i - 24 i + 12 (i^{1} i)$

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

$a + b i = 90 - 45 i - 24 i + 12 (i^{1} i^{1})$

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

$a + b i = 90 - 45 i - 24 i + 12 i^{1 + 1}$

Add $1$ and $1$ .

$a + b i = 90 - 45 i - 24 i + 12 i^{2}$

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

$a + b i = 90 - 45 i - 24 i + 12 \cdot - 1$

Multiply $12$ by $- 1$ .

$a + b i = 90 - 45 i - 24 i - 12$

Subtract $12$ from $90$ .

$a + b i = 78 - 45 i - 24 i$

Subtract $24 i$ from $- 45 i$ .

$a + b i = 78 - 69 i$

Step 3

Subtract $b i$ from both sides of the equation.

$a = 78 - 69 i - b i$

Solve for a (15-4i)(6-3i)=a+bi

Related Questions

Support LCMGCF.com by disabling your adblocker.