First slide

Algebra of complex numbers

Question

$If m and x are two real numbers and i = \sqrt{- 1}, then e^{2 m i \cot^{- 1} x} {(\frac{x i + 1}{x i - 1})}^{m} =$

Moderate

A

2
B

0
C

1
D

3

Solution

$Let {Cot}^{- 1} x = θ, then x = \cot θ$
$∴ e^{2 m i \cot^{- 1} x} {(\frac{x i + 1}{x i - 1})}^{m} = e^{2 m i θ} {(\frac{i \cot θ + 1}{i \cot θ - 1})}^{m}$
$= e^{2 m i θ} {(\frac{i (\cot θ - i)}{i (\cot θ + i)})}^{m} = e^{2 m i θ} {(\frac{\cos θ - i \sin θ}{\cos θ + i \sin θ})}^{m}$
$= e^{2 m i θ} {(\frac{e^{- i θ}}{e^{i θ}})}^{m} = e^{2 m i θ} e^{- 2 i θ m} = 1$

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