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MHT CET · Maths · Complex Number

The value of \(\frac{\mathrm{i}^{248}+\mathrm{i}^{246}+\mathrm{i}^{244}+\mathrm{i}^{242}+\mathrm{i}^{240}}{\mathrm{i}^{249}+\mathrm{i}^{247}+\mathrm{i}^{245}+\mathrm{i}^{243}+\mathrm{i}^{241}}\), \((\mathrm{i}=\sqrt{-1})\) is

  1. A \(\mathrm{i}\)
  2. B 1
  3. C -1
  4. D \(-\mathrm{i}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(-\mathrm{i}\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \frac{i^{245}+i^{246}+i^{244}+i^{242}+i^{240}}{i^{243}+i^{247}+i^{245}+i^{243}+i^{241}} \\ & =\frac{i^{240}\left(i^8+i^6+i^4+i^2+1\right)}{i^{241}\left(i^8+i^6+i^4+i^2+1\right)} \\ & =\frac{i^{200}}{i^{241}} \\ & =\frac{1}{i} \\ & =\frac{i}{i^2}=-i\end{aligned}\)