Question

If $z{\left(2-i2\sqrt{3}\right)}^2=i{\left(\sqrt{3}+i\right)}^4$, then amplitude of $z$ is

• A. $-\frac{\mathrm{\pi }}{6}$
• B. $\frac{\mathrm{\pi }}{4}$
• C. $\frac{\mathrm{\pi }}{6}$
• D. None of these

Answer 1

The correct option is A

$-\frac{\mathrm{\pi }}{6}$

Explanation for the correct option.

Step 1. Simplify the equation and solve for $z$.

Let $\omega$ be the cube root of unit. Therefore, ${\omega }^3=1$ and $1+\omega +{\omega }^2=1$.

Where $\omega =\frac{-1+i\sqrt{3}}{2}$ and ${\omega }^2=\frac{-1-i\sqrt{3}}{2}$.

Now, in the given equation $z{\left(2-i2\sqrt{3}\right)}^2=i{\left(\sqrt{3}+i\right)}^4$, divide both sides by $16$ and write in terms of $\omega$.

$$\begin{gathered} \frac{z{\left(2-i2\sqrt{3}\right)}^2}{16}=\frac{i{\left(\sqrt{3}+i\right)}^4}{16} \\ \Rightarrow z{\left(\frac{2-i2\sqrt{3}}{4}\right)}^2=i{\left(\frac{\sqrt{3}+i}{2}\right)}^4 \\ \Rightarrow z{\left(\frac{1-i\sqrt{3}}{2}\right)}^2=i{\left(\frac{1}{i}\frac{\sqrt{3}i+i^2}{2}\right)}^4 \\ \Rightarrow z{\left(\frac{-1+i\sqrt{3}}{2}\right)}^2=i\times \frac{1}{i^4}{\left(\frac{-1+i\sqrt{3}}{2}\right)}^4 \\ \Rightarrow z{\omega }^2=i{\omega }^4\left[\omega =\frac{-1+i\sqrt{3}}{2},i^4=1\right] \\ \Rightarrow z=i{\omega }^2 \end{gathered}$$

Step 2. Find the amplitude of $z$.

In the equation $z=i{\omega }^2$, substitute ${\omega }^2=\frac{-1-i\sqrt{3}}{2}$.

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

So, the point $z=\frac{\sqrt{3}}{2}-\frac{1}{2}i$ lies in the third quadrant, so its argument is given as:

$\begin{array}{rcl}\arg \left(z\right)& =& {\tan }^{-1}\left(\frac{-{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{\sqrt{3}}{2}}}\right)\\ & =& {\tan }^{-1}\left(\frac{-1}{\sqrt{3}}\right)\\ & =& -\frac{\mathrm{\pi }}{6}\end{array}$

So, the amplitude of $z$ is $-\frac{\mathrm{\pi }}{6}$.

Hence, the correct option is A.