The problem is to simplify the complex fraction $\frac{1 + \sqrt{2}i}{\sqrt{3} - 2i}$.

1. Problem Description

The problem is to simplify the complex fraction $\frac{1 + \sqrt{2}i}{\sqrt{3} - 2i}$.

2. Solution Steps

To simplify the complex fraction, we multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $\sqrt{3} - 2i$ is $\sqrt{3} + 2i$.

\begin{align*}

\frac{1 + \sqrt{2}i}{\sqrt{3} - 2i} &= \frac{1 + \sqrt{2}i}{\sqrt{3} - 2i} \cdot \frac{\sqrt{3} + 2i}{\sqrt{3} + 2i} \\

&= \frac{(1 + \sqrt{2}i)(\sqrt{3} + 2i)}{(\sqrt{3} - 2i)(\sqrt{3} + 2i)}

\end{align*}

Now, we multiply the terms in the numerator and denominator.

Numerator:

\begin{align*}

(1 + \sqrt{2}i)(\sqrt{3} + 2i) &= 1(\sqrt{3}) + 1(2i) + \sqrt{2}i(\sqrt{3}) + \sqrt{2}i(2i) \\

&= \sqrt{3} + 2i + \sqrt{6}i + 2\sqrt{2}i^2 \\

&= \sqrt{3} + 2i + \sqrt{6}i - 2\sqrt{2} \\

&= (\sqrt{3} - 2\sqrt{2}) + (2 + \sqrt{6})i

\end{align*}

Denominator:

\begin{align*}

(\sqrt{3} - 2i)(\sqrt{3} + 2i) &= (\sqrt{3})^2 - (2i)^2 \\

&= 3 - 4i^2 \\

&= 3 - 4(-1) \\

&= 3 + 4 \\

&= 7

\end{align*}

So,

\begin{align*}

\frac{1 + \sqrt{2}i}{\sqrt{3} - 2i} &= \frac{(\sqrt{3} - 2\sqrt{2}) + (2 + \sqrt{6})i}{7} \\

&= \frac{\sqrt{3} - 2\sqrt{2}}{7} + \frac{2 + \sqrt{6}}{7}i

\end{align*}

3. Final Answer

The simplified form of the given complex fraction is $\frac{\sqrt{3} - 2\sqrt{2}}{7} + \frac{2 + \sqrt{6}}{7}i$.