Question

Find the square of $ \sqrt{9+40\sqrt{-1}}+\sqrt{9-40\sqrt{-1}} $ .

Answer 1

Hint: In order to solve this problem, we need to first solve the inner square root. We can do this by assuming $ \sqrt{-1}=i $ , this makes the question less confusing and now need to deal with only one square-root sign. In this question, we need to use the identity $ {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab $ . Also, we need to know that $ {{i}^{2}}=-1 $ .

Complete step-by-step answer:
We need to find the square of the above expression.
Before that we need to write $ \sqrt{-1}=i $ , $ i $ is a complex number. This will make the calculation easier.
Substituting $ \sqrt{-1}=i $ we get,
 $ \sqrt{9+40i}+\sqrt{9-40i} $ .
Squaring the above equation, we get,
 $ {{\left( \sqrt{9+40i}+\sqrt{9-40i} \right)}^{2}}={{\left( \sqrt{9+40i} \right)}^{2}}+{{\left( \sqrt{9-40i} \right)}^{2}}+2\left( \sqrt{9+40i} \right)\left( \sqrt{9-40i} \right) $
We have used the identity that $ {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab $
Simplifying we get,
The square and the square-root neutralize the effect giving just the following expression,
 $ \begin{align}
  & {{\left( \sqrt{9+40i}+\sqrt{9-40i} \right)}^{2}}=\left( 9+40i \right)+\left( 9-40i \right)+2\left( \sqrt{9+40i} \right)\left( \sqrt{9-40i} \right) \\
 & =18+2\left[ \sqrt{\left( 9\times 9 \right)+\left( 9\times -40i \right)+\left( 9\times 40i \right)+\left( 40i\times -40i \right)} \right]
\end{align} $
Simplifying further, and substituting $ {{i}^{2}}=-1 $ , we get,
 $ \begin{align}
  & {{\left( \sqrt{9+40i}+\sqrt{9-40i} \right)}^{2}}=18+2\left[ \sqrt{81+{{40}^{2}}} \right] \\
 & =18+2\left[ \sqrt{81+1600} \right] \\
 & =18+2\sqrt{1681} \\
 & =18+\left( 2\times 41 \right) \\
 & =18+82 \\
 & =100
\end{align} $

Note: As we can see that all the complex numbers “I” have been cancelled out and we remain with the natural numbers. Also, it is easily missed that we need to find the square rather than directly simplifying the expression. The negative sign helps to cancel the cross-terms containing all the complex numbers.