Question

Find the magnitude and amplitude of the respective expression $\dfrac{{\left( {1 + i\sqrt 3 } \right)\left( {2 + 2i} \right)}}{{\sqrt 3 - i}}$
(a) $2\sqrt 2$ and $\dfrac{\pi }{4}$ respectively
(b) $\dfrac{\pi }{4}$ and $2\sqrt 2$ respectively
(c) Cannot be determined
(d) None of these

Answer 1

Hint : The given problem revolves around the concepts of solving complex relations that is $z = a + bi$ ‘ $a$ ’ is ‘real part’ and ‘ $b$ ’ is the imaginary part ‘ $i$ ’ mathematically represented as ‘ $\sqrt { - 1}$ ’. First of all take conjugate to the given expression i.e. $\sqrt 3 + i$ (opposite to denominator). As a result, comparing the solution with the complex equation substituting the respective values in magnitude $\sqrt {{a^2} + {b^2}}$ and amplitude $\theta = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)$ so as to get the desire value.

Complete step-by-step answer :
The given expression is $\dfrac{{\left( {1 + i\sqrt 3 } \right)\left( {2 + 2i} \right)}}{{\sqrt 3 - i}}$
As the expression contains an imaginary unit that is ‘ $i$ ’,
So, equating ‘ $z$ ’ to the respective expression, we get
$z = \dfrac{{\left( {1 + i\sqrt 3 } \right)\left( {2 + 2i} \right)}}{{\sqrt 3 - i}}$
To find the magnitude and amplitude, simplifying the equation in simplest form,
So, multiplying and dividing the equation by its conjugate that is $\sqrt 3 + i$ , we get
$z = \dfrac{{2 + 2i + 2\sqrt 3 i + 2\sqrt 3 {i^2}}}{{\sqrt 3 - i}} \times \dfrac{{\sqrt 3 + i}}{{\sqrt 3 + i}}$
We know that $i = \sqrt { - 1} \to {i^2} = - 1 \to {i^3} = - i$ , as a result, substituting it, we get $z = \dfrac{{\left( {2 + 2i + 2\sqrt 3 i - 2\sqrt 3 } \right)}}{{\sqrt 3 - i}} \times \dfrac{{\sqrt 3 + i}}{{\sqrt 3 + i}}$
Separating the real part and imaginary part, we get

$$z = \dfrac{{\left( {\sqrt 3 + i} \right)\left[ {\left( {2 - 2\sqrt 3 } \right) + \left( {2i + 2i\sqrt 3 } \right)} \right] }}{{\left( {\sqrt 3 - i} \right)\left( {\sqrt 3 + i} \right)}} = \dfrac{{\left( {\sqrt 3 + i} \right)\left( {2 - 2\sqrt 3 } \right) + \left( {2i + 2i\sqrt 3 } \right)\left( {\sqrt 3 + i} \right)}}{{{{\left( {\sqrt 3 } \right)}^2} - {i^2}}} \\\ \Rightarrow z = \dfrac{{2\sqrt 3 - 6 + 2i - 2\sqrt 3 i + 2\sqrt 3 i + 2{i^2} + 6i + 2\sqrt 3 {i^2}}}{{3 + 1}}\;$$

Solving the equation mathematically, we get

$$z = \dfrac{{ - 6 + 2i - 2 + 6i}}{4} \\\ \Rightarrow z = \dfrac{{ - 8 + 8i}}{4} \;$$

Since, multiplying and dividing the equation by $4$ , we get
$z = - 2 + 2i$
Now, since Comparing with the complex relations that is $z = a + bi$ , we get
Magnitude of the expression or equation seems to be $= \sqrt {{a^2} + {b^2}}$
Hence, substituting the values in the equation, we get
$\therefore$ Magnitude $= \sqrt {{2^2} + {{\left( { - 2} \right)}^2}} = \sqrt 8 = 2\sqrt 2$
Similarly,
For Amplitude of the expression or equation seems to be, $\theta = {\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)$ , we get
Amplitude, $\tan \theta = \dfrac{2}{{\left( { - 2} \right)}} = \left( { - 1} \right) \Rightarrow \theta = {\tan ^{ - 1}}\left( { - 1} \right)$
We know that, $\tan {45^ \circ } = \tan \dfrac{\pi }{4} = 1{\text{ also, }}\tan \left( { - \theta } \right) = - \tan \theta$
The equation becomes,
$\therefore$ Amplitude $= \theta = - {\tan ^{ - 1}}\left( 1 \right) = {45^ \circ } = \dfrac{\pi }{4}$ respectively
$\therefore$ magnitude and amplitude of the given expression $\dfrac{{\left( {1 + i\sqrt 3 } \right)\left( {2 + 2i} \right)}}{{\sqrt 3 - i}}$ is $2\sqrt 2$ and $\dfrac{\pi }{4}$ respectively. So, option (a) is correct.

Note : One must be able to know the basic idea behind these complex relations. Then, comparing the desired solutions with the desired general complex equation i.e. $z = a + bi$ , the required parameters can be solved, so as to be sure of our final answer. Algebraic knowledge leads to the ease of the problem as per the complexity of the problem.