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TS EAMCET · Maths · Complex Number

The roots of the equation \((x-1)^5=32(x+1)^5\) are

  1. A \(\frac{1+2 e^{\frac{2 k \pi i}{5}}}{1-2 e^{\frac{2 k \pi i}{5}}}, k=1,2,3,4,5\)
  2. B \(\frac{1-2 e^{\frac{2 k \pi i}{5}}}{1+2 e^{\frac{2 k \pi i}{5}}}, k=0,1,2,3,4\)
  3. C \(1,2 \omega, 3 \omega^2, 2 \omega+3 \omega^2, 5 \omega^2+7\)
  4. D \(\frac{3+2 e^{\frac{2(k+1) \pi i}{5}}}{3-2 e^{\frac{2(k+1) \pi i}{5}}}, k=0,1,2,3,4\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1+2 e^{\frac{2 k \pi i}{5}}}{1-2 e^{\frac{2 k \pi i}{5}}}, k=1,2,3,4,5\)

Step-by-step Solution

Detailed explanation

We have, \((x-1)^5=32(x+1)^5\) \(\Rightarrow\left(\frac{x-1}{x+1}\right)^5=32 \Rightarrow \frac{x-1}{x+1}=(32)^{1 / 5} \Rightarrow \frac{x-1}{x+1}=2 e^{\frac{2 k \pi i}{5}}\) Apply componendo and dividendo…