Question

divide the following complex numbers and express the result in standard form, ( a + bi ), where ( a ) and ( b ) are fractions in lowest terms. (\frac{1 + 3i}{3 + 3i}) (\frac{1 + 3i}{3 + 3i}=square)

Explanation:

Step1: Multiply by conjugate of denominator

Multiply numerator and denominator by $3-3i$ (the conjugate of $3+3i$):
$$\frac{1+3i}{3+3i} \times \frac{3-3i}{3-3i} = \frac{(1+3i)(3-3i)}{(3+3i)(3-3i)}$$

Step2: Expand numerator using FOIL

Calculate product of numerator terms:
$$(1)(3) + (1)(-3i) + (3i)(3) + (3i)(-3i) = 3 - 3i + 9i - 9i^2$$
Substitute $i^2=-1$:
$$3 + 6i - 9(-1) = 3 + 6i + 9 = 12 + 6i$$

Step3: Expand denominator (difference of squares)

Use $(a+b)(a-b)=a^2-b^2$ and $i^2=-1$:
$$3^2 - (3i)^2 = 9 - 9i^2 = 9 - 9(-1) = 9 + 9 = 18$$

Step4: Simplify the fraction

Split the fraction and reduce to lowest terms:
$$\frac{12 + 6i}{18} = \frac{12}{18} + \frac{6}{18}i = \frac{2}{3} + \frac{1}{3}i$$

Answer:

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