Question

Find the modulus and amplitude for each of the following complex numbers.
$\left( i \right)7-5i$
$\left( ii \right)\sqrt{3}+\sqrt{2i}$
$\left( iii \right)-8+15i$
$\left( iv \right)-3\left( 1-i \right)$
$\left( v \right)-4-4i$
$\left( vi \right)\sqrt{3}-i$
$\left( vii \right)3$
$\left( viii \right)1+i$
$\left( ix \right)1+i\sqrt{3}$
$\left( x \right){{\left( 1+2i \right)}^{2}}\left( 1-i \right)$

Answer 1

Hint : To find the modulus of the given complex numbers, we will use the formula, that for a complex number, $x+iy$ , the modulus will be $\sqrt{{{x}^{2}}+{{y}^{2}}}$ and its amplitude will be ${{\tan }^{-1}}\left( \dfrac{y}{x} \right)$ . By using these formulas, we can get the required answers.

Complete step-by-step answer :
In the question, we are given a set of complex numbers and have been asked to find their modulus and amplitude. Let us know a bit about complex numbers before we start solving. A complex number is a number that can be written in the form of $a+bi$ , where a, b are the real numbers and I is a solution of the equation ${{x}^{2}}=-1$ . This is because no real number satisfies the equation ${{x}^{2}}+1=0\Rightarrow {{x}^{2}}=-1$ , hence it is called an imaginary number. For the complex number $a+ib$ , a is considered as the real part and b as the imaginary part. Despite the historical nomenclature, imaginary complex numbers are regarded in mathematical science as real as any real number and are fundamental in any aspect of scientific description of the natural world. We find the modulus and amplitude by using the formula, $\sqrt{{{x}^{2}}+{{y}^{2}}}$ and ${{\tan }^{-1}}\left( \dfrac{y}{x} \right)$ , if the given complex number is in the form of $x+iy$ . So, we will find it for each of the given numbers.
$\left( i \right)7-5i$
Here the complex number is $7-5i$ , so the value of x and y are 7 and - 5 respectively. So,
Modulus = $\sqrt{{{7}^{2}}+{{\left( -5 \right)}^{2}}}\Rightarrow \sqrt{49+25}\Rightarrow \sqrt{74}$ .
Amplitude = ${{\tan }^{-1}}\left( \dfrac{-5}{7} \right)$ .
$\left( ii \right)\sqrt{3}+\sqrt{2i}$
Here, we have the value of x and y as $\sqrt{3}$ and $\sqrt{2}$ respectively. So,
Modulus = $\sqrt{{{\left( \sqrt{3} \right)}^{2}}+{{\left( \sqrt{2} \right)}^{2}}}\Rightarrow \sqrt{3+2}\Rightarrow \sqrt{5}$ .
Amplitude = ${{\tan }^{-1}}\left( \dfrac{\sqrt{2}}{\sqrt{3}} \right)\Rightarrow {{\tan }^{-1}}\left( \sqrt{\dfrac{2}{3}} \right)$ .
$\left( iii \right)-8+15i$
Here, the value of x and y are given as - 8 and 15 respectively. So, we can say,
Modulus = $\sqrt{{{\left( -8 \right)}^{2}}+{{15}^{2}}}\Rightarrow \sqrt{64+225}\Rightarrow \sqrt{289}\Rightarrow 17$ .
Amplitude = ${{\tan }^{-1}}\left( \dfrac{15}{-8} \right)\Rightarrow {{\tan }^{-1}}\left( \dfrac{-15}{8} \right)$ .
$\left( iv \right)-3\left( 1-i \right)$
We can simplify this given complex number as $-3+3i$ . So, we will get the values of x and y as - 3 and 3 respectively. So, we can calculate,
Modulus = $\sqrt{{{\left( -3 \right)}^{2}}+{{3}^{2}}}\Rightarrow \sqrt{9+9}\Rightarrow \sqrt{18}\Rightarrow 3\sqrt{2}$ .
Amplitude = ${{\tan }^{-1}}\left( \dfrac{3}{-3} \right)\Rightarrow {{\tan }^{-1}}\left( -1 \right)\Rightarrow -{{\tan }^{-1}}\left( 1 \right)$ , which can be written as $-\dfrac{\pi }{4}$ .
$\left( v \right)-4-4i$
Now, we have the values of x and y as - 4 and - 4 here. So, the values of,
Modulus = $\sqrt{{{\left( -4 \right)}^{2}}+{{\left( -4 \right)}^{2}}}\Rightarrow \sqrt{16+16}\Rightarrow \sqrt{32}\Rightarrow 4\sqrt{2}$ .
Amplitude = ${{\tan }^{-1}}\left( \dfrac{-4}{-4} \right)\Rightarrow {{\tan }^{-1}}\left( 1 \right)$ , which can be written as $\dfrac{\pi }{4}$ .
$\left( vi \right)\sqrt{3}-i$
Now, here we have the value of x and y as $\sqrt{3}$ and -1 respectively. So, we get,
Modulus = $\sqrt{{{\left( \sqrt{3} \right)}^{2}}+{{\left( -1 \right)}^{2}}}\Rightarrow \sqrt{3+1}\Rightarrow \sqrt{4}\Rightarrow 2$ .
Amplitude = ${{\tan }^{-1}}\left( \dfrac{-1}{\sqrt{3}} \right)\Rightarrow -{{\tan }^{-1}}\left( \dfrac{1}{\sqrt{3}} \right)$ , which can be written as $-\dfrac{\pi }{3}$ .
$\left( vii \right)3$
We can write 3 as $3+0i$ also. So, in that case, the values of x and y will be 3 and 0. So,
Modulus = $\sqrt{{{3}^{2}}+{{0}^{2}}}\Rightarrow \sqrt{9}\Rightarrow 3$ .
Amplitude = ${{\tan }^{-1}}\left( \dfrac{0}{3} \right)\Rightarrow {{\tan }^{-1}}\left( 0 \right)\Rightarrow 0{}^\circ$ .
$\left( viii \right)1+i$
So, here we have the values of x and y as 1 and 1. So, we get,
Modulus = $\sqrt{{{1}^{2}}+{{1}^{2}}}\Rightarrow \sqrt{1+1}\Rightarrow \sqrt{2}$ .
Amplitude = ${{\tan }^{-1}}\left( \dfrac{1}{1} \right)\Rightarrow {{\tan }^{-1}}\left( 1 \right)$ , which can be written as $\dfrac{\pi }{4}$ .
$\left( ix \right)1+i\sqrt{3}$
The values of x and y here are 1 and $\sqrt{3}$ , so we get,
Modulus = $\sqrt{{{1}^{2}}+{{\left( \sqrt{3} \right)}^{2}}}\Rightarrow \sqrt{1+3}\Rightarrow \sqrt{4}\Rightarrow 2$ .
Amplitude = ${{\tan }^{-1}}\left( \dfrac{\sqrt{3}}{1} \right)\Rightarrow {{\tan }^{-1}}\left( \sqrt{3} \right)$ , which can be written as $\dfrac{\pi }{3}$ .
$\left( x \right){{\left( 1+2i \right)}^{2}}\left( 1-i \right)$
To solve this, we have to first simplify it. We will use the identity, ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ for the same. So, we get,
$\left( 1+4{{i}^{2}}+4i \right)\left( 1-i \right)$
We know that ${{i}^{2}}=-1$ , so by substituting it, we get,
$\begin{aligned} & \left( 1-4+4i \right)\left( 1-i \right) \\\ & \Rightarrow \left( -3+4i \right)\left( 1-i \right) \\\ & \Rightarrow -3+3i+4i-4{{i}^{2}} \\\ \end{aligned}$
Now, again substituting the value of ${{i}^{2}}=-1$ , we get,
$-3+7i+4\Rightarrow \left( 1+7i \right)$ .
So, we get the values of x and y as 1 and 7. So, we can get the,
Modulus = $\sqrt{{{1}^{2}}+{{7}^{2}}}\Rightarrow \sqrt{1+49}\Rightarrow \sqrt{50}\Rightarrow 5\sqrt{2}$ .
Amplitude = ${{\tan }^{-1}}\left( 7 \right)$ .

Note : For a given complex number, its modulus represents its distance from the origin and its amplitude represents the angle subtended by the line joining the complex number and the origin with x - axis. The possible mistakes that the students can make is by writing the value of ${{i}^{2}}$ as 1, but this is wrong and the students must note that ${{i}^{2}}=-1$ .