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

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Answer:

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

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

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