Question

Find the square root of the complex number - 4 – 3i
a. $\pm \dfrac{1}{\sqrt{2}}\left( 1-3i \right)$
b. $\pm \dfrac{1}{\sqrt{3}}\left( 3+i \right)$
c. $\pm \dfrac{1}{\sqrt{3}}\left( 3+2i \right)$
d. $\pm \dfrac{1}{\sqrt{3}}\left( 4+i \right)$

Answer 1

Hint: In order to solve this question, we will consider the square root of complex as a + ib, because the square root of a complex number is also a complex number. Also, we should know that ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$ and ${{\left( i \right)}^{2}}=-1$. By using these, we will compare the result and get the answer.

Complete step-by-step answer:
In this question, we have been asked to find the square root of a complex number, (- 4 – 3i). To solve this question, let us consider (a + ib) as the square root of (- 4 – 3i). So, we can write,
${{\left( a+ib \right)}^{2}}=-4-3i$
Now, we know that ${{\left( x+y \right)}^{2}}={{x}^{2}}+{{y}^{2}}+2xy$. So, for x = a and y = ib, we can write,
${{\left( a+ib \right)}^{2}}={{a}^{2}}+{{\left( ib \right)}^{2}}+2a\left( ib \right)=-4-3i$
We can further write it as,
${{a}^{2}}+{{\left( i \right)}^{2}}{{b}^{2}}+2abi=-4-3i$
Now, we know that ${{\left( i \right)}^{2}}=-1$. So, we can write,
${{a}^{2}}-{{b}^{2}}+2abi=-4-3i$
Now, we will compare real and complex parts of both side of the equations, so we can write,
${{a}^{2}}-{{b}^{2}}=-4$ and 2ab = -3
${{a}^{2}}-{{b}^{2}}=-4.........\left( i \right)$ and $a=\dfrac{-3}{2b}.........\left( ii \right)$
Now, we will put the value of a from equation (ii) in equation (i), so we will get,
$\begin{aligned} & {{\left( \dfrac{-3}{2b} \right)}^{2}}-{{b}^{2}}=-4 \\\ & \dfrac{9}{4{{b}^{2}}}-{{b}^{2}}=-4 \\\ & \dfrac{9-4{{b}^{4}}}{4{{b}^{2}}}=-4 \\\ & 9-4{{b}^{4}}=-16{{b}^{2}} \\\ & 4{{b}^{4}}-16{{b}^{2}}-9=0 \\\ \end{aligned}$
By applying the middle term splitting method, we will get,
$\begin{aligned} & 4{{b}^{4}}-18{{b}^{2}}+2{{b}^{2}}-9=0 \\\ & 2{{b}^{2}}\left( 2{{b}^{2}}-9 \right)+1\left( 2{{b}^{2}}-9 \right)=0 \\\ & \left( 2{{b}^{2}}-9 \right)\left( 2{{b}^{2}}+1 \right)=0 \\\ & 2{{b}^{2}}-9=0;2{{b}^{2}}+1=0 \\\ & {{b}^{2}}=\dfrac{9}{2};{{b}^{2}}=-\dfrac{1}{2} \\\ & b=\pm \dfrac{3}{\sqrt{2}};b=\pm \dfrac{1}{\sqrt{2}}i \\\ \end{aligned}$
Now, we will put the values of b in equation (ii). So, we get,
For $b=\pm \dfrac{3}{\sqrt{2}}$ we get, $a=\dfrac{-3}{2\left( \pm \dfrac{3}{\sqrt{2}} \right)}=\mp \dfrac{1}{\sqrt{2}}$
For $b=\pm \dfrac{1}{\sqrt{2}}i$ we get, $a=\dfrac{-3}{2\left( \pm \dfrac{1}{\sqrt{2}}i \right)}=\mp \dfrac{3}{\sqrt{2}i}=\pm \dfrac{3}{\sqrt{2}}i$
So, we can write, (a + ib) as,
For $a=\pm \dfrac{1}{\sqrt{2}}$ and $b=\mp \dfrac{3}{\sqrt{2}}i$,
$a+ib=\pm \dfrac{1}{\sqrt{2}}\mp \dfrac{3}{\sqrt{2}}i=\pm \dfrac{1}{\sqrt{2}}\left( 1-3i \right)$
And for $a=\pm \dfrac{3}{\sqrt{2}}i$ and $b=\pm \dfrac{1}{\sqrt{2}}i$,
$a+ib=\pm \dfrac{3}{\sqrt{2}}i\pm \dfrac{1}{\sqrt{2}}{{i}^{2}}=\pm \dfrac{3}{\sqrt{2}}i\mp \dfrac{1}{\sqrt{2}}=\pm \dfrac{1}{\sqrt{2}}\left( 1-3i \right)$
In both the cases we get the value of (a + ib) as $\pm \dfrac{1}{\sqrt{2}}\left( 1-3i \right)$. Therefore, we can say that the correct square root of (- 4 – 3i) is $\pm \dfrac{1}{\sqrt{2}}\left( 1-3i \right)$.
Hence, option (a) is the correct answer.

Note: While solving this question, the possible mistake we can make is by writing $b=\pm \dfrac{3}{\sqrt{2}}i$ for $a=\pm \dfrac{1}{2}$ which is wrong and will give us the wrong answer. We can also solve this question by hit and trial method, by finding out the square of every option and finding the correct answer, but it will be a very time consuming process. Hence, the preferred method is the conventional method.