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 $i^2=-1$:

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