### Represent the complex number $z = \sqrt{ 3 } + i$ in the polar form.

$2 \left( cos \dfrac{ \pi } { 6 } + i \space sin \dfrac{ \pi } { 6 } \right)$

Step by Step Explanation:
1. We have, \begin{align} & z = \sqrt{ 3 } + i \\ \end{align} The standard polar form of a complex number is $r(cos \theta + i \space sin \theta)$
2. On comparing $z$ with the standard polar form of a complex number, we get,
$r \space cos \space \theta = \sqrt{ 3 }$ and $r \space sin \space \theta = 1$
Now, \begin{align} & r \space cos \space \theta = \sqrt{ 3 } && \ldots (1) \\ \implies & r^2 \space cos^2 \theta = \sqrt{ 3 } ^2 && \ldots (2) \\ & r \space sin \theta = 1 && \ldots (3) \\ \implies & r^2 \space sin^2 \theta = 1 ^2 && \ldots (4) \end{align} On Adding $(2)$ and $(4)$ we get,
\begin{align} & r^2 \space cos^2 \theta + r^2 \space sin^2 \theta = \sqrt{ 3 } ^2 + 1 ^2 \\ \implies & r^2 ( cos^2 \theta + sin^2 \theta ) = 3 + 1 \\ \implies & r^2 = 4 && [\text{Since, } cos^2 \theta + sin^2 \theta = 1] \\ \implies & r = 2 && [\text{Conventionally } r > 0] \end{align}
3. Substituting the value of $r$ in eq $(1)$ and $(3)$ we get,
$cos \theta = \dfrac{ \sqrt{ 3 } }{ 2 }$ and $sin \theta = \dfrac{ 1 }{ 2 }$
$\implies \theta = \dfrac{ \pi } { 6 }$
4. Hence, the polar form of the complex number $z = \sqrt{ 3 } + i$ is $2 \left( cos \dfrac{ \pi } { 6 } + i \space sin \dfrac{ \pi } { 6 } \right)$.