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Properties of Conjugate of a Complex Number

If α is a r...

Question

If $α$ is a root of $x^{4} = 1$ which negative principle argument, then the principle argument of $△ (α) =$ $∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 α^{n} & α^{n + 1} & α^{n + 3} \frac{1}{α^{n + 1}} & \frac{1}{α^{n}} & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣$ is

A

\frac{5 π}{14}

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B

\frac{3 π}{4}

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C

\frac{π}{4}

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D

- \frac{π}{4}

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Solution

The correct option is A $\frac{5 π}{14}$
Given determinante is not right

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^{'} α^{'}

x^{4} = 1

with negative principal argument, then the principal argument of

Δ (α)

Δ (α) = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 α^{n} & α^{n + 1} & α^{n + 3} \frac{1}{α^{n + 1}} & \frac{1}{α^{n}} & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣

y (α) = \sqrt{2 (\frac{tan α + cot α}{1 + {tan}^{2} α}) + \frac{1}{{sin}^{2} α}}

α \in (\frac{3 π}{4}, π)

\frac{d y}{d α}

α = \frac{5 π}{6}

y (α) = \sqrt{2 (\frac{tan α + cot α}{1 + {tan}^{2} α}) + \frac{1}{{sin}^{2} α}}

α \in (\frac{3 π}{4}, π)

\frac{d y}{d α}

α = \frac{5 π}{6}

Which of the following statements are correct?

1 . if sin θ = sin α ⇒ θ = nπ + $(- 1)^{n}$ α where α ∈ [ - $\frac{π}{2}$ , $\frac{π}{2}$ ] n ∈ I

2 . if cos θ = cos α ⇒ 2nπ±α where α ∈ [0,π] n ∈ I

3 . if tan θ = tan α ⇒ θ = nπ + α where α ∈ (- $\frac{π}{2}$ , $\frac{π}{2}$ ) n ∈ I

α

β

are the roots of

x^{2} - 2 x + 4 = 0

α^{n} - β^{n} = i 2^{n + 1} sin \frac{n π}{3}

α^{9} - β^{9}

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