Which of the following complex numbers is equivalent to \(\dfrac{2+3i}{4-2i} \) ?

(Note: \(i=\sqrt{-1})\)

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Notice how all of the denominators of the answer choices do not contain \(i\). Similar to radicals, we need to rationalize the denominator. We can do this by multiplying the numerator and denominator by the conjugate of the denominator. In other words:

$$ \frac{2+3i}{4-2i} \cdot \frac{4+2i}{4+2i} $$ $$ =\frac{8+12i+4i+6i^2}{16-4i^2} $$

Since \(i^2=-1\):

$$ \frac{8+16i-6}{20} $$ $$ \frac{2+16i}{20} $$ $$ = \frac{2}{20}+\frac{16i}{20} $$ $$ = \boxed{\frac{1}{10}+\frac{4i}{5}} $$