Brazil
Convert the complex number ^@ \dfrac{ 4 }{ 1 - \sqrt{ 3 } i } ^@ into polar form.
Answer:
^@ 2 \left( cos \dfrac{ \pi } { 3 } + i \space sin \dfrac{ \pi } { 3 } \right) ^@
- ^@ \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} ^@
- 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) ^@
- 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} @^ - 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 } ^@ - 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) ^@.
