Question

If $z=\frac{1+i\sqrt{3}}{\sqrt{3}+i}$, then ${\overline{z}}^{100}$ lies in the

• A. first quadrant
• B. second quadrant
• C. third quadrant
• D. fourth quadrant

Answer 1

The correct option is C

third quadrant

Explanation for the correct option.

Step 1. Write $z$ in euler's form

For the complex number $z=\frac{1+i\sqrt{3}}{\sqrt{3}+i}$ multiply the numerator and denominator by $\sqrt{3}-i$.

$\begin{array}{rcl}z& =& \frac{\left(1+i\sqrt{3}\right)\left(\sqrt{3}-i\right)}{\left(\sqrt{3}+i\right)\left(\sqrt{3}-i\right)}\\ & =& \frac{\sqrt{3}-i+3i-i^2\sqrt{3}}{{\left(\sqrt{3}\right)}^2-i^2}\\ & =& \frac{\sqrt{3}+2i-(-1)\sqrt{3}}{3-(-1)}\left[i^2=-1\right]\\ & =& \frac{\sqrt{3}+2i+\sqrt{3}}{3+1}\\ & =& \frac{2\sqrt{3}+2i}{4}\\ & =& \frac{\sqrt{3}}{2}+\frac{1}{2}i\end{array}$

Now, as $\cos \frac{\mathrm{\pi }}{6}=\frac{\sqrt{3}}{2}$ and $\sin \frac{\mathrm{\pi }}{6}=\frac{1}{2}$, so $z$ can be written in euler's form as:

$\begin{array}{rcl}z& =& \frac{\sqrt{3}}{2}+\frac{1}{2}i\\ & =& \cos \frac{\mathrm{\pi }}{6}+i\sin \frac{\mathrm{\pi }}{6}\\ & =& e^{i\frac{\mathrm{\pi }}{6}}\left[e^{i\theta }=\cos \theta +i\sin \theta \right]\end{array}$

Step 2. Find the value of $\overline{z}$.

For a complex number $z=e^{i\theta }$, the value of the conjugate is given as:

$\overline{z}=e^{i\left(-\theta \right)}$.

So the conjugate of $z=\begin{array}{rcl}& & e\end{array}\begin{array}{rcl}& & \end{array}$ is $\overline{z}=e^{i\left(-\frac{\mathrm{\pi }}{6}\right)}$.

Step 3. Find the quadrant in which ${\overline{z}}^{100}$ lies.

As $\overline{z}=e^{i\left(-\frac{\mathrm{\pi }}{6}\right)}$, so ${\overline{z}}^{100}$ is given as:

$\begin{array}{rcl}{\overline{z}}^{100}& =& {\left(e^{i\left(-\frac{\mathrm{\pi }}{6}\right)}\right)}^{100}\\ & =& e^{i\left(-\frac{\mathrm{\pi }}{6}\times 100\right)}\\ & =& e^{i\left(-\frac{50\mathrm{\pi }}{3}\right)}\end{array}$

Now, expand $\begin{array}{rcl}& & e^{i\left(-\frac{50\mathrm{\pi }}{3}\right)}\end{array}$ using euler's form as:

$\begin{array}{rcl}{e}^{i\left(-\frac{50\mathrm{\pi }}{3}\right)}& =& \cos \left(-\frac{50\mathrm{\pi }}{3}\right)+i\sin \left(-\frac{50\mathrm{\pi }}{3}\right)\left[e^{i\theta }=\cos \theta +i\sin \theta \right]\\ & =& \cos \left(\frac{50\mathrm{\pi }}{3}\right)-i\sin \left(\frac{50\mathrm{\pi }}{3}\right)\left[\cos \left(-\theta \right)=\cos \theta ,\sin \left(-\theta \right)=-\sin \theta \right]\\ & =& \cos \left(16\mathrm{\pi }+\frac{2\mathrm{\pi }}{3}\right)-i\sin \left(16\mathrm{\pi }+\frac{2\mathrm{\pi }}{3}\right)\\ & =& \cos \left(\frac{2\mathrm{\pi }}{3}\right)-i\sin \left(\frac{2\mathrm{\pi }}{3}\right)\left[\cos \left(2n\mathrm{\pi }+\mathrm{\theta }\right)=\cos \theta ,\sin \left(2n\mathrm{\pi }+\mathrm{\theta }\right)=\sin \theta \right]\\ & =& -\frac{1}{2}-i\frac{\sqrt{3}}{2}\left[\cos \left(\frac{2\mathrm{\pi }}{3}\right)=-\frac{1}{2},\sin \left(\frac{2\mathrm{\pi }}{3}\right)=\frac{\sqrt{3}}{2}\right]\end{array}$

So the complex number ${\overline{z}}^{100}$ is equal to $\begin{array}{rcl}& & -\frac{1}{2}-i\frac{\sqrt{3}}{2}\end{array}$.

As both its real and imaginary parts are negative.

So the complex number ${\overline{z}}^{100}$ lies in the third quadrant.

Hence, the correct option is C.