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Find the value of: $\dfrac{{{i^6} + {i^7} + {i^8} + {i^9}}}{{{i^2} + {i^3}}}$

Answer
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Hint: Use the information ${i^2} = - 1$ and ${i^4} = 1$.

We know that, ${i^2} = - 1$ and ${i^4} = 1$. Using these formulae, we can write these imaginary numbers as:
$
  {i^6} = {i^4} \times {i^2} = 1 \times - 1 = - 1 \\
  {i^7} = {i^1} \times {i^4} \times {i^2} = i \times 1 \times - 1 = - i \\
  {i^8} = {i^4} \times {i^4} = 1 \times 1 = 1 \\
  {i^9} = {i^1} \times {i^4} \times {i^4} = i \times 1 \times 1 = i \\
  {i^3} = {i^2} \times i = - i \\
$

Using above information in the given question,
$\dfrac{{{i^6} + {i^7} + {i^8} + {i^9}}}{{{i^2} + {i^3}}} \Rightarrow \dfrac{{ - 1 - i + 1 + i}}{{ - 1 - i}} \Rightarrow \dfrac{0}{{ - 1 - i}} = 0$
Hence, the value of the given expression is 0.

Note: There are n number of questions which depend upon the same concept. We just need to calculate the expression in powers of i and then putting the value will give us the answer.
Last updated date: 27th Sep 2023
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