Question

Find the square root of the complex number – 21 – 20i.
(a) $\pm \left( 2+5i \right)$
(b) $\pm \left( 2-5i \right)$
(c) $\pm \left( 4+3i \right)$
(d) $\pm \left( 4-3i \right)$

Answer 1

Hint: We will take the help of the formula for finding square roots of a quadratic equation of the form $a{{x}^{2}}+bx+c=0$. This formula is written as $x=\dfrac{-{{b}_{1}}\pm \sqrt{{{b}_{1}}^{2}-4{{a}_{1}}c}}{2{{a}_{1}}}$. We will solve this question by taking the square root of the number in the form of $a+ib$. Therefore, we will write $\sqrt{-21-20i}=a+ib$. By these tricks we will be able to solve the question correctly.

Complete step-by-step answer:
We will consider the complex number – 21 – 20i and suppose that the square root of this number is in the form of $a+ib$. Therefore, we have $\sqrt{-21-20i}=a+ib$. Now we will square this equation so that we can get rid of the root. Therefore, after squaring both the sides we will have ${{\left( \sqrt{-21-20i} \right)}^{2}}={{\left( a+ib \right)}^{2}}$.
Now we will apply the formula of one of the algebraic formulas. The one is given by ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$. Thus the value of ${{\left( a+ib \right)}^{2}}={{a}^{2}}+{{\left( ib \right)}^{2}}+2aib$. Since, ${{\left( \sqrt{-21-20i} \right)}^{2}}={{\left( a+ib \right)}^{2}}$ or $-21-20i={{\left( a+ib \right)}^{2}}$ thus, we have $-21-20i={{a}^{2}}+{{i}^{2}}{{b}^{2}}+2aib$.
As we know that the value of ${{i}^{2}}=-1$ therefore, after substituting the value in the equation $-21-20i={{a}^{2}}+{{i}^{2}}{{b}^{2}}+2aib$ we will get
$\begin{aligned} & -21-20i={{a}^{2}}+\left( -1 \right){{b}^{2}}+2aib \\\ & \Rightarrow -21-20i={{a}^{2}}-{{b}^{2}}+2aib \\\ \end{aligned}$
After comparing the constants and the i terms on both the sides of the equation we will have ${{a}^{2}}-{{b}^{2}}=-21$ and $2ab=-20$. By considering the equation $2ab=-20$ we will take the value of b in terms of a. This is done as $b=\dfrac{-20}{2a}$ or $b=\dfrac{-10}{a}$. Now we will substitute the value of b in equation ${{a}^{2}}-{{b}^{2}}=-21$. Thus, we get
$\begin{aligned} & {{a}^{2}}-{{\left( \dfrac{-10}{a} \right)}^{2}}=-21 \\\ & \Rightarrow {{a}^{2}}-\dfrac{100}{{{a}^{2}}}=-21 \\\ \end{aligned}$
By the lcm we will now have
$\begin{aligned} & \dfrac{{{a}^{4}}-100}{{{a}^{2}}}=-21 \\\ & \Rightarrow {{a}^{4}}-100=-21{{a}^{2}} \\\ & \Rightarrow {{a}^{4}}+21{{a}^{2}}-100=0 \\\ \end{aligned}$
Now we will put x as ${{a}^{2}}$. Therefore, our equation changes into ${{x}^{2}}+21x-100=0$. Now we will apply the formula for finding the roots of the quadratic equation. The formula is given by $x=\dfrac{-{{b}_{1}}\pm \sqrt{{{b}_{1}}^{2}-4{{a}_{1}}c}}{2{{a}_{1}}}$. By keeping ${{a}_{1}}$ as 1, ${{b}_{1}}$ as 21 and c = - 100 we will get,
$\begin{aligned} & x=\dfrac{-21\pm \sqrt{{{\left( 21 \right)}^{2}}-4\left( 1 \right)\left( -100 \right)}}{2\left( 1 \right)} \\\ & \Rightarrow x=\dfrac{-21\pm \sqrt{441+400}}{2} \\\ & \Rightarrow x=\dfrac{-21\pm \sqrt{841}}{2} \\\ & \Rightarrow x=\dfrac{-21\pm 29}{2} \\\ \end{aligned}$
Therefore, we get two roots and these are given by $x=\dfrac{-21+29}{2}$ or $x=\dfrac{-21-29}{2}$. After simplifying these we will get $x=\dfrac{8}{2}$ or $x=\dfrac{-50}{2}$. Thus, we now have that x = 4 or x = -25. As we know that x is as ${{a}^{2}}$ thus we get $x={{a}^{2}}=4$ which results in a = $\pm$ 2. Similarly, $x={{a}^{2}}=25$ which results in $x=\pm \sqrt{25}$.
We will neglect $x=\pm \sqrt{25}$ and consider a = $\pm$ 2. Now, we will put a = 2 in the equation $b=\dfrac{-10}{2}$ which results into b = - 5. As we know that the square root of this number is in the form of $a+ib$. Therefore, we have $\sqrt{-21-20i}=a+ib$ which is actually equal to $\pm$ (2 – 5i).
Hence, the correct option is (b).

Note: One can mistake during the substitution in, $x={{a}^{2}}=25$ as one might forget to put ${{a}^{2}}$ in x and solve further. This will lead to wrong answers. Whenever we are performing with the roots of any number then we should not forget to write it along with both plus and minus signs. That is a = $\pm$ 2 is going to be considered instead of a = 2. We can also use a hit and trial method for solving the quadratic equation. This can be done by substituting the integers one by one and collecting the first number which satisfies the quadratic equation ${{x}^{2}}+21x-100=0$.