Question

Find $\sqrt{5-12i}$?
(a) $3-2i$
(b) $3+2i$
(c) \[3-i\]
(d) $4-2i$

Answer 1

Hint: Assume the given expression as E. Now, write $12i=2\times 6\times i$, where $i$ is an imaginary number given as $i=\sqrt{-1}$, and compare it with $2\times a\times b$ to find the values of a and b. In the next step, split 5 into two terms such that it becomes of the form ${{a}^{2}}+{{b}^{2}}$. Finally, use the algebraic identity ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$ and then use the formula of exponent given as ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ to get the answer.

Complete step by step solution:
Here we have been provided with the expression $\sqrt{5-12i}$ and we are asked to find its value, that means we need to find the square root of the expression $\left( 5-12i \right)$. Let us assume this expression as E, so we have,
$\Rightarrow E=\left( 5-12i \right)$
Now, we need to convert it in the form ${{a}^{2}}+{{b}^{2}}-2ab$ so that we can apply the algebraic identity and get the answer. In the expression ${{a}^{2}}+{{b}^{2}}-2ab$ we have three terms and in the given expression $\left( 5-12i \right)$ we have only two terms. So we need to split one of the terms. When we compare the two expressions then we see that before $12i$ we have minus sign so we need to compare it with 2ab and choose the value of a and b such that the expression \[{{a}^{2}}+{{b}^{2}}\] results in 5. So on comparing $12i$ with 2ab we get,
\[\begin{align}
  & \Rightarrow 12i=2\times a\times b \\
 & \Rightarrow 2\times 6\times i=2\times 3\times 2i \\
\end{align}\]
From the above relation we can assume $a=3$ and $b=2i$, so squaring and adding a and b we get,
\[\begin{align}
  & \Rightarrow {{a}^{2}}+{{b}^{2}}={{3}^{2}}+{{\left( 2i \right)}^{2}} \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}=9+4{{i}^{2}} \\
\end{align}\]
Here $i$ is an imaginary number given as $i=\sqrt{-1}$, so substituting this value in the above expression we get,
\[\begin{align}
  & \Rightarrow {{a}^{2}}+{{b}^{2}}=9+4\left( -1 \right) \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}=9-4 \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}=5 \\
\end{align}\]
So we can write $\left( 5-12i \right)$ in the form ${{a}^{2}}+{{b}^{2}}-2ab$ as:
\[\begin{align}
  & \Rightarrow \left( 5-12i \right)={{3}^{2}}+{{\left( 2i \right)}^{2}}-2\left( 3 \right)\left( 2i \right) \\
 & \Rightarrow E={{3}^{2}}+{{\left( 2i \right)}^{2}}-2\left( 3 \right)\left( 2i \right) \\
\end{align}\]
Using the algebraic identity ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$ we get,
\[\Rightarrow E={{\left( 3-2i \right)}^{2}}\]
Taking square root both the sides we get,
\[\begin{align}
  & \Rightarrow \sqrt{E}=\sqrt{5-12i} \\
 & \Rightarrow \sqrt{E}=\sqrt{{{\left( 3-2i \right)}^{2}}} \\
 & \Rightarrow \sqrt{E}=\sqrt{{{\left( 3-2i \right)}^{2}}} \\
\end{align}\]
Using the formula of exponent ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ we get,
\[\Rightarrow \sqrt{E}=\left( 3-2i \right)\]

So, the correct answer is “Option A”.

Note: Here you must not consider $i$ as any variable because if you will do so then you may not be able to break 5 into \[{{a}^{2}}+{{b}^{2}}\]. The expression $i$ is called iota and it is the solution of the quadratic equation ${{x}^{2}}+1=0$. In case you find difficulty in choosing a and b then you can also form a bi – quadratic equation and solve for the two values using the relation $2ab=12i$ and ${{a}^{2}}+{{b}^{2}}=5$.