The roots of the equation (x-1)^5=32(x+1)^5 a | vedclass

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(A) Given the equation $(x-1)^5=32(x+1)^5$. Dividing both sides by $(x+1)^5$,we get $\left(\frac{x-1}{x+1}\right)^5=32$. Let $z = \frac{x-1}{x+1}$. Th

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$\frac{1+2 e^{\frac{2 k \pi i}{5}}}{1-2 e^{\frac{2 k \pi i}{5}}}, k=0,1,2,3,4$

Vedclass · Answer

(A) Given the equation $(x-1)^5=32(x+1)^5$. Dividing both sides by $(x+1)^5$,we get $\left(\frac{x-1}{x+1}\right)^5=32$. Let $z = \frac{x-1}{x+1}$. Then $z^5 = 32 = 2^5 \cdot e^{i(2k\pi)}$,where $k=0, 1, 2, 3, 4$. Thus,$z = 2 e^{\frac{2k\pi i}{5}}$. Substituting back,$\frac{x-1}{x+1} = 2 e^{\frac{2k\pi i}{5}}$. Let $\alpha = 2 e^{\frac{2k\pi i}{5}}$. Then $\frac{x-1}{x+1} = \alpha$. $x-1 = \alpha(x+1)$ $\Rightarrow x(1-\alpha) = 1+\alpha$ $\Rightarrow x = \frac{1+\alpha}{1-\alpha}$. Substituting $\alpha$ back,$x = \frac{1+2 e^{\frac{2k\pi i}{5}}}{1-2 e^{\frac{2k\pi i}{5}}}$ for $k=0, 1, 2, 3, 4$.

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

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