Given-z-6-2-3-i-find-the-two-least-value-of-n-such-that-z-n-z-purelly-imaginary-

Question Number 112737 by bemath last updated on 09/Sep/20

Given z = - 6 + 2 \sqrt{3} i

find the two least value of n such

that z^{n} . \bar{z} purelly imaginary .

Commented by bemath last updated on 09/Sep/20

thank you both sir

Answered by mathmax by abdo last updated on 09/Sep/20

z = - 6 + 2 \sqrt{3} i \Rightarrow∣ z ∣ = \sqrt{36 + 12} = \sqrt{48} = 4 \sqrt{3} \Rightarrow z = 4 \sqrt{3} {\frac{- 6}{4 \sqrt{3}} + \frac{2 \sqrt{3}}{4 \sqrt{3}} i}

= 4 \sqrt{3} {\frac{- 3}{2 \sqrt{3}} + \frac{1}{2} i} = 4 \sqrt{3} {- \frac{\sqrt{3}}{2} + \frac{1}{2} i} = 4 \sqrt{3} e^{i (\frac{5 π}{6})} \Rightarrow

z^{n} = {(4 \sqrt{3})}^{n} e^{i \frac{5 n π}{6}} \Rightarrow z^{n} . \bar{z} = {(4 \sqrt{3})}^{n} e^{i \frac{5 n π}{6}} . 4 \sqrt{3} e^{- i \frac{5 π}{6}}

= {(4 \sqrt{3})}^{n + 1} e^{i (\frac{5}{6} (n - 1)) π} = {(4 \sqrt{3})}^{n + 1} {\cos (\frac{5 (n - 1) π}{6}) + isin (\frac{5 (n - 1) π}{6})}

z^{n} . \bar{z} \in iR \Leftrightarrow \cos (\frac{5 (n - 1) π}{6}) = 0 \Leftrightarrow \frac{5 (n - 1) π}{6} = \frac{π}{2} + k π \Rightarrow

\frac{5 n - 5}{6} = \frac{1}{2} + k \Rightarrow 5 n - 5 = 3 + 6 k \Rightarrow 5 n = 8 + 6 k \Rightarrow n = \frac{6 k + 8}{5} we search k /

5 divide 6 k + 8 \dots .

Answered by john santu last updated on 09/Sep/20

(1) z = - 6 + 2 \sqrt{3} i = {r e}^{i θ}

\to r = \sqrt{36 + 12} = \sqrt{48} = 4 \sqrt{3}

\to \tan θ = \frac{2 \sqrt{3}}{- 6} = \frac{\sqrt{3}}{- 3}, θ = \frac{5 π}{6}

z = 4 \sqrt{3} e^{\frac{5 π}{6} i}

(2) \bar{z} = - 6 - 2 \sqrt{3} i = {r e}^{i α}

\to \tan α = \frac{- 2 \sqrt{3}}{- 6} = \frac{\sqrt{3}}{3}, α = \frac{7 π}{6}

\bar{z} = 4 \sqrt{3} e^{\frac{7 π}{6} i}

(3) z^{n} . \bar{z} = {(4 \sqrt{3} e^{\frac{5 π}{6} i})}^{n} (4 \sqrt{3} e^{\frac{7 π}{6} i}) = 0 + b i

\to {(4 \sqrt{3})}^{n + 1} (\cos (\frac{5 π n}{6} + \frac{7 π}{6}) + i \sin (\frac{5 π n}{6} + \frac{7 π}{6})) = 0 + b i

w e g e t \cos (\frac{(5 n + 7) π}{6}) = 0

\to \cos (\frac{(5 n + 7) π}{2}) = \cos (\frac{π}{2})

\to \frac{(5 n + 7) π}{6} = \pm \frac{π}{2} + m . 2 π

\to (5 n + 7) π = \pm 3 π + k . 2 π

\to {\begin{cases} 5 n + 7 = 3 + 2 k, 5 n = - 4 + 2 k \\ 5 n + 7 = - 3 + 2 k, 5 n = - 10 + 2 k \end{cases}

\to 5 n = - 10 + 2 k {\begin{cases} k = 10, n = 2 \\ k = 15, n = 4 \end{cases}