Question

The magnitude and amplitude of $\frac{(1+i\sqrt{3})(2+2i)}{\left(\sqrt{3}-i\right)}$ are respectively

• A. $2,\frac{3\mathrm{\pi }}{4}$
• B. $2\sqrt{2},\frac{3\mathrm{\pi }}{4}$
• C. $2\sqrt{2},\frac{\mathrm{\pi }}{4}$
• D. $2\sqrt{2},\frac{\mathrm{\pi }}{2}$

Answer 1

The correct option is C

$2\sqrt{2},\frac{\mathrm{\pi }}{4}$

Explanation of the correct option.

Step $1$: Simplify the fraction.

Given : $\frac{(1+i\sqrt{3})(2+2i)}{\left(\sqrt{3}-i\right)}$

Multiply the numerator and denominator with $\left(\sqrt{3}+i\right)$,

$$\begin{gathered} \frac{(1+i\sqrt{3})(2+2i)\times \left(\sqrt{3}+i\right)}{\left(\sqrt{3}-i\right)\times \left(\sqrt{3}+i\right)} \\ \Rightarrow \frac{2(1+i)(\sqrt{3}+i+i3-\sqrt{3})}{3+1} \\ \Rightarrow \frac{2(1+i)\left(i4\right)}{4} \\ \Rightarrow \frac{4i-4}{2} \\ \Rightarrow -2+2i \end{gathered}$$

Step $2$: Compute the magnitude.

Since magnitude of $x+iy$ is given by $\left|x+iy\right|=\sqrt{x^2+y^2}$,

$$\begin{gathered} \left|-2+2i\right|=\sqrt{4+4} \\ \Rightarrow \left|-2+2i\right|=2\sqrt{2} \end{gathered}$$

Step $3$: Compute the amplitude.

Since amplitude of $x+iy$ is given by $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$,

$$\begin{gathered} \theta ={\tan }^{-1}\frac{2}{-2} \\ \Rightarrow \theta ={\tan }^{-1}\left(-1\right) \\ \Rightarrow \theta =\frac{\mathrm{\pi }}{4} \end{gathered}$$

Therefore, magnitude and amplitude of $\frac{(1+i\sqrt{3})(2+2i)}{\left(\sqrt{3}-i\right)}$ is $2\sqrt{2}$ and $\frac{\mathrm{\pi }}{4}$ respectively.

Hence option $\left(\mathrm{C}\right)$ is the correct option.