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Convert the complex number $\dfrac{ 4 }{ 1 - \sqrt{ 3 } i }$ into polar form.

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Answer:

$2 \left( cos \dfrac{ \pi } { 3 } + i \space sin \dfrac{ \pi } { 3 } \right)$

Step by Step Explanation:

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

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