soit α∈]0;π[. determiner: 1)le module et l′argument de: a)1−e^(iα) ,b)1+e^(i𝛂) 2)deduire le module et l′argument de a) ((1−e^(iα) )/(1+e^(iα) )), b)(1−e^(iα) )(1+e^(iα) ) rochinel930@gmail.c


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Question Number 81954 by Cmr 237 last updated on 16/Feb/20

$s o i t α \in] 0; π [. d e t e r m i n e r :$ $1) l e m o d u l e e t l^{'} a r g u m e n t d e :$ $a) 1 - e^{i α}, b) 1 + e^{i α}$ $2) d e d u i r e l e m o d u l e e t l^{'} a r g u m e n t d e$ $a) \frac{1 - e^{i α}}{1 + e^{i α}}, b) (1 - e^{i α}) (1 + e^{i α})$ $r o c h i n e l 930 @ g m a i l . c$
Commented by abdomathmax last updated on 16/Feb/20
$z=1−e^(iα) =1−cosα−isinα =2sin^2 ((α/2))−2isin((α/2))cos((α/2)) =2sin((α/2))(sin((α/2))−icos((α/2))) 0<(α/2)<(π/2) ⇒sin((α/2))>0 ⇒z=2sin((α/2)){cos((π/2)−(α/2))−isin((π/2)−(α/2))} =2sin((α/2)){cos(((α−π)/2))+isin(((α−π)/2))} ⇒ ∣z∣=2sin((α/2)) and arg(z)=((α−π)/2) [2π]$
$z = 1 - e^{i α} = 1 - c o s α - i s i n α = 2 {s i n}^{2} (\frac{α}{2}) - 2 i s i n (\frac{α}{2}) c o s (\frac{α}{2})$ $= 2 s i n (\frac{α}{2}) (s i n (\frac{α}{2}) - i c o s (\frac{α}{2}))$ $0 < \frac{α}{2} < \frac{π}{2} \Rightarrow s i n (\frac{α}{2}) > 0 \Rightarrow z = 2 s i n (\frac{α}{2}) {c o s (\frac{π}{2} - \frac{α}{2}) - i s i n (\frac{π}{2} - \frac{α}{2})}$ $= 2 s i n (\frac{α}{2}) {c o s (\frac{α - π}{2}) + i s i n (\frac{α - π}{2})} \Rightarrow$ $∣ z ∣= 2 s i n (\frac{α}{2}) a n d a r g (z) = \frac{α - π}{2} [2 π]$
Commented by abdomathmax last updated on 16/Feb/20
$Z=1+e^(iα) ⇒Z=1+cosα +isinα =2cos^2 ((α/2))+2isin((α/2))cos((α/2)) =2cos((α/2)){cos((α/2))+isin((α/2))} cos((α/2))>0 due to 0<α<π ⇒∣Z∣=2cos((α/2))and argZ=(α/2)[2π]$
$Z = 1 + e^{i α} \Rightarrow Z = 1 + c o s α + i s i n α$ $= 2 {c o s}^{2} (\frac{α}{2}) + 2 i s i n (\frac{α}{2}) c o s (\frac{α}{2})$ $= 2 c o s (\frac{α}{2}) {c o s (\frac{α}{2}) + i s i n (\frac{α}{2})}$ $c o s (\frac{α}{2}) > 0 d u e t o 0 < α < π \Rightarrow∣ Z ∣= 2 c o s (\frac{α}{2}) a n d$ $a r g Z = \frac{α}{2} [2 π]$
Commented by abdomathmax last updated on 16/Feb/20

$∣ \frac{1 - e^{i α}}{1 + e^{i α}} ∣ = \frac{∣ 1 - e^{i α} ∣}{∣ 1 + e^{i α} ∣} = \frac{2 s i n (\frac{α}{2})}{2 c o s (\frac{α}{2})} = t a n (\frac{α}{2}) a n d$ $a r g (\frac{1 - e^{i α}}{1 + e^{i α}}) = a r g (1 - e^{i α}) - a r g (1 + e^{i α}) [2 π]$ $= \frac{α - π}{2} - \frac{α}{2} [2 π] = - \frac{π}{2} [2 π]$
Commented by abdomathmax last updated on 16/Feb/20
$∣(1−e^(iα) )(1+e^(iα) )∣=2sin((α/2))×2cos((α/2)) =2sin(α) and arg{(1−e^(iα) )(1+e^(iα) )} =arg(1−e^(iα) ) +arg(1+e^(iα) )[2π] =((α−π)/2) +(α/2) [2π] =α−(π/2)$
$∣ (1 - e^{i α}) (1 + e^{i α}) ∣= 2 s i n (\frac{α}{2}) \times 2 c o s (\frac{α}{2})$ $= 2 s i n (α) a n d a r g {(1 - e^{i α}) (1 + e^{i α})}$ $= a r g (1 - e^{i α}) + a r g (1 + e^{i α}) [2 π]$ $= \frac{α - π}{2} + \frac{α}{2} [2 π] = α - \frac{π}{2}$
Commented by Cmr 237 last updated on 16/Feb/20

$1) 1 + e^{i α} = e^{\frac{i α}{2}} (e^{- \frac{i α}{2}} + e^{\frac{i α}{2}})$ $= 2 c o s (\frac{α}{2}) e^{\frac{i α}{2}}$ $∣ 1 + e^{i α} ∣=∣ 2 c o s (\frac{α}{2}) e^{\frac{i α}{2}} ∣$ $= 2 c o s (\frac{α}{2})$ $∣ 1 - e^{i α} ∣=∣ - 2 i s i n (\frac{α}{2}) e^{\frac{i α}{2}} ∣$ $= 2 s i n (\frac{α}{2})$ $a r g (1 - e^{i α}) = a r g (- 2 i s i n (\frac{α}{2}) e^{\frac{i α}{2}})$ $= a r g (- 2 s i n (\frac{α}{2})) + a r g (i) + a r g (e^{\frac{i α}{2}})$ $= π + \frac{π}{2} + \frac{α}{2} + 2 k π$ $= \frac{3 π + α}{2} + 2 k π$
Commented by mathmax by abdo last updated on 16/Feb/20

$\frac{3 π + α}{2} + 2 k π = 2 π - \frac{π}{2} + \frac{α}{2} + 2 k π = \frac{α - π}{2} [2 π]$
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