If \[z = \dfrac{{1 + i\sqrt 3 }}{{\sqrt 3 + i}}\],then \[{(\mathop z\limits^ - )^{100}}\] lies in which quadrant.
A) First quadrant.
B) Second quadrant.
C) Third quadrant.
D) Fourth quadrant.

Hint:
To find the quadrant of the given \[{(\mathop z\limits^ - )^{100}}\], firstly we have to simplify the given value of \[z\] and take a conjugate of the simplified value.
After taking a conjugate we have to take a power of \[z\] as \[100\], which is already given in the question. By using De Moivre’s Theorem we resolve the power of \[100\] and after solving it we get the quadrant of \[{(\mathop z\limits^ - )^{100}}\].

Formula used:
The formula we used here is shown below:
1. \[z = r\left( {\cos \theta + i\sin \theta } \right)\] where \[r\] is \[\sqrt {{a^2} + {b^2}} \] and \[\theta \] is \[{\tan ^{ - 1}}\left( {\dfrac{{{\mathop{\rm Im}\nolimits} \left( z \right)}}{{{\mathop{\rm Re}\nolimits} \left( z \right)}}} \right)\].
2. \[{\left( {\cos \theta + i\sin \theta } \right)^n} = \left( {\cos n\theta + i\sin n\theta } \right)\]
3. \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\]

Complete step-by-step solution:
The given equation is
\[z = \dfrac{{1 + i\sqrt 3 }}{{\sqrt 3 + i}}\]
Firstly, we will multiply numerator and denominator of the given function with \[\sqrt 3 - i\], we get
 \[z = \dfrac{{1 + i\sqrt 3 }}{{\sqrt 3 + i}} \times \dfrac{{\sqrt 3 - i}}{{\sqrt 3 - i}}\]
Now, we will use this identity \[\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}\] in the denominator, we get
\[z = \dfrac{{\sqrt 3 - i + 3i - \sqrt 3 {i^2}}}{{{{(\sqrt 3 )}^2} - {{(i)}^2}}}\]
As we know that \[{i^2} = - 1\].
So, further we will substitute this value in the above equation, we get
\[\begin{array}{l}z = \dfrac{{\sqrt 3 + 2i - \left( { - \sqrt 3 } \right)}}{{3 - ( - 1)}}\\z = \dfrac{{2\sqrt 3 + 2i}}{4}\end{array}\]
Furthermore, we will rewrite the above equation as
\[z = \dfrac{{\sqrt 3 }}{2} + \dfrac{i}{2}\]
Now, we compare the above equation with the formula \[z = r\left( {\cos \theta + i\sin \theta } \right)\] where where \[r\] is \[\sqrt {{a^2} + {b^2}} \] and \[\theta \] is \[{\tan ^{ - 1}}\left( {\dfrac{{{\mathop{\rm Im}\nolimits} \left( z \right)}}{{{\mathop{\rm Re}\nolimits} \left( z \right)}}} \right)\].
Here, \[{\mathop{\rm Im}\nolimits} \left( z \right)\] is \[\dfrac{1}{2}\] and \[{\mathop{\rm Re}\nolimits} \left( z \right) = \dfrac{{\sqrt 3 }}{2}\].
Now, we will find \[\theta \] and \[r\], we get
\[\begin{array}{l}r = \sqrt {{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2} + {{\left( {\dfrac{1}{2}} \right)}^2}} \\r = \sqrt {\dfrac{3}{4} + \dfrac{1}{4}} \\r = 1\end{array}\]
And
\[\begin{array}{l}\theta = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}}} \right)\\\theta = {\tan ^{ - 1}}\sqrt 3 \end{array}\]
As we know that \[\tan \dfrac{\pi }{6} = \sqrt 3 \].
So, we get
\[\begin{array}{l}\theta = {\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{6}} \right)\\\theta = \dfrac{\pi }{6}\end{array}\]
Further, we will substitute these values in the formula \[z = r\left( {\cos \theta + i\sin \theta } \right)\], we get
\[z = \cos \dfrac{\pi }{6} + i\sin \dfrac{\pi }{6}\]
Further, we will find \[{(\mathop z\limits^ - )^{100}}\].
To find \[{(\mathop z\limits^ - )^{100}}\] firstly we will take the conjugate of \[z\]
As we know, the conjugate of the real function cannot change the sign whereas the conjugate of the imaginary function changes the sign from positive to negative.
Therefore, we get,
\[\overline z = \cos \left( {\dfrac{\pi }{6}} \right) - i\sin \left( {\dfrac{\pi }{6}} \right)\]
Now we will take\[100\] as a power of conjugate of term \[z\], we get;
\[{\left( {\mathop z\limits^ - } \right)^{100}} = {\left[ {\cos \left( {\dfrac{\pi }{6}} \right) - i\sin \left( {\dfrac{\pi }{6}} \right)} \right]^{100}}\]
Further, we will apply the demoivre’s theorem that is \[{\left( {\cos \theta + i\sin \theta } \right)^n} = \left( {\cos n\theta + i\sin n\theta } \right)\], we get
\[\begin{array}{l}{\left( {\mathop z\limits^ - } \right)^{100}} = \cos \left( {\dfrac{{100\pi }}{6}} \right) - i\sin \left( {\dfrac{{100\pi }}{6}} \right)\\{\left( {\mathop z\limits^ - } \right)^{100}} = \cos \left( {\dfrac{{50\pi }}{3}} \right) - i\sin \left( {\dfrac{{50\pi }}{3}} \right)\end{array}\]
Now, we simplify \[\theta \] in the above expression in the multiple of \[2\pi \] to find the quadrant, we get
\[\begin{array}{l}{\left( {\mathop z\limits^ - } \right)^{100}} = \cos \left( {16\pi + \dfrac{{2\pi }}{3}} \right) - i\sin \left( {16\pi + \dfrac{{2\pi }}{3}} \right)\\{\left( {\mathop z\limits^ - } \right)^{100}} = \cos \left( {\dfrac{{2\pi }}{3}} \right) - i\sin \left( {\dfrac{{2\pi }}{3}} \right)\end{array}\]
Further, we will again simplifying the \[\theta \] in the above expression, we get
\[{\left( {\mathop z\limits^ - } \right)^{100}} = - \cos \left( {\dfrac{\pi }{3}} \right) - i\sin \left( {\dfrac{\pi }{3}} \right)\]
Now we can see that the \[\sin \]and \[\cos \]is negative. Therefore, it is only the third quadrant in which both are negative.

Hence, Option C) is the correct answer.

Note:
In this type of question, we should remember how to convert a complex number equation into a trigonometric equation and we should always remember how to find conjugate of any complex number and also Demoivre’s theorem. We should be very careful while simplifying the trigonometric function values and also remember the concept of a quadrant.