Question

If an expression is given as ${{\left( \sqrt{8}+i \right)}^{50}}={{3}^{49}}\left( a+ib \right),$ then find the value of ${{a}^{2}}+{{b}^{2}}$ .
$\begin{align}
  & \text{A}\text{. }\left( {{a}^{2}}+{{b}^{2}} \right)=9 \\
 & \text{B}\text{. }\left( {{a}^{2}}+{{b}^{2}} \right)=27 \\
 & \text{C}\text{. }\left( {{a}^{2}}+{{b}^{2}} \right)=3 \\
 & \text{D}\text{. }\left( {{a}^{2}}+{{b}^{2}} \right)=1 \\
\end{align}$

Answer 1

Hint: Here, the question is given in complex numbers which consist of real & imaginary numbers. To find this first we need to take the imaginary number as $i=\sqrt{-1}$ and substitute it in the equation given in the question i.e. ${{\left( \sqrt{8}+i \right)}^{50}}={{3}^{49}}\left( a+ib \right),$ and simplify it to get the value of ${{a}^{2}}+{{b}^{2}}$.

Complete step by step solution:
A complex number is a combination of real and imaginary numbers. We know that imaginary number such that $i=\sqrt{-1}$.
By squaring on both sides, we get –
${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}$
${{i}^{2}}=-1$
Modulus value of -1 is $\left| -1 \right|=+1$
$\therefore {{i}^{2}}=1$
In this question we are given that ${{\left( \sqrt{8}+i \right)}^{50}}={{3}^{49}}\left( a+ib \right).$
We know that modulus of complex numbers $\left( a+ib \right)=\sqrt{{{a}^{2}}+{{b}^{2}}}$ .
Here, $\left( \sqrt{8}+i \right)$and $\left( a+ib \right)$ are complex numbers.
By taking modulus of complex numbers, we get –
\[{{\left( \sqrt{{{\left( \sqrt{8} \right)}^{2}}+{{\left( i \right)}^{2}}} \right)}^{50}}={{3}^{49}}\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right)\]
By simplifying the above equations, we get –
${{\left( \sqrt{8+{{i}^{2}}} \right)}^{50}}={{3}^{49}}\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right)$
By squaring on both sides in the above equation, we get –
${{\left( {{\left( \sqrt{8+{{i}^{2}}} \right)}^{50}} \right)}^{2}}={{\left( {{3}^{49}}\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right) \right)}^{2}}$
By simplifying the above equation, we get –
\[{{\left( {{\left( \sqrt{8+{{i}^{2}}} \right)}^{2}} \right)}^{50}}={{\left( {{3}^{49}} \right)}^{2}}{{\left( \sqrt{{{a}^{2}}+{{b}^{2}}} \right)}^{2}}\]
By cancelling the roots and squares, we get –
\[{{\left( 8+{{i}^{2}} \right)}^{50}}={{\left( {{3}^{49}} \right)}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right).\]
Here, we will substitute ${{i}^{2}}=1$ in the above equation, so we get –
${{\left( 8+1 \right)}^{50}}={{\left( {{3}^{49}} \right)}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)$
${{\left( 9 \right)}^{50}}={{\left( {{3}^{49}} \right)}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)$
We know that ${{3}^{2}}=9$ .
By taking ${{3}^{2}}$ instead of 9 we get –
${{\left( {{3}^{2}} \right)}^{50}}={{\left( {{3}^{49}} \right)}^{2}}\left( {{a}^{2}}+{{b}^{2}} \right)$
We know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ . So, we get –
${{3}^{100}}={{3}^{98}}\left( {{a}^{2}}+{{b}^{2}} \right)$
By dividing ${{3}^{98}}$ on both sides, we get –
$\dfrac{{{3}^{100}}}{{{3}^{98}}}=\left( {{a}^{2}}+{{b}^{2}} \right)$
We know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ , so we get –
${{3}^{2}}={{a}^{2}}+{{b}^{2}}$
$9={{a}^{2}}+{{b}^{2}}$
Therefore, ${{a}^{2}}+{{b}^{2}}=9$ .
Hence, option (A) is the correct answer.

Note: Students should be very careful while simplifying the equation. We need ${{a}^{2}}+{{b}^{2}}$to be at one side of the equation to find an answer. Students may also find this question with the help of formula given in De Moivre’s theorem i.e. ${{\left( \cos \theta +i\sin \theta \right)}^{n}}=\left( \cos n\theta +i\sin n\theta \right)$ but this will be the lengthy procedure to find the solution. So, we will ignore it.