Question

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which of the following complex numbers is equivalent to \\(\frac{3 - 5i}{8 + 2i}\\)? (note: \\(i = \sqrt{-1}\\)) a) \\(\frac{3}{8} - \frac{5i}{2}\\) b) \\(\frac{3}{8} + \frac{5i}{2}\\) c) \\(\frac{7}{34} - \frac{23i}{34}\\) d) \\(\frac{7}{34} + \frac{23i}{34}\\)

Explanation:

Step1: Multiply by conjugate of denominator

Multiply numerator and denominator by $8-2i$:
$$\frac{3-5i}{8+2i} \times \frac{8-2i}{8-2i} = \frac{(3-5i)(8-2i)}{(8+2i)(8-2i)}$$

Step2: Expand numerator using FOIL

Calculate product of binomials:
$$(3)(8) + (3)(-2i) + (-5i)(8) + (-5i)(-2i) = 24 -6i -40i +10i^2$$
Substitute $i^2=-1$:
$$24 -46i +10(-1) = 24-10 -46i = 14 -46i$$

Step3: Expand denominator (difference of squares)

Use $a^2 - (bi)^2 = a^2 + b^2$:
$$8^2 - (2i)^2 = 64 -4i^2 = 64 -4(-1) = 64+4=68$$

Step4: Simplify the fraction

Divide numerator and denominator by 2:
$$\frac{14-46i}{68} = \frac{7}{34} - \frac{23i}{34}$$

Answer:

C) $\frac{7}{34} - \frac{23i}{34}$