1) Aspet Algébrique
On désigne par i \bf i i imaginaire tel que i 2 = − 1 i^2 =-1 i 2 = − 1 nombre complexe tout nombre z z z
où a et b sont des nombres réels. Cette écriture est appelée forme alg e ˋ brique {\bf\text{forme algèbrique}} forme alg e ˋ brique z z z a a a b b b partie r e ˊ elle {\bf \text{partie réelle}} partie r e ˊ elle partie imaginaire {\bf \text{partie imaginaire}} partie imaginaire z z z
a = R e ( z ) e t b = I m ( z ) a=\mathcal{R}e(z)~~~ et~~~ b=\mathcal{I}m(z) a = R e ( z ) e t b = I m ( z ) On désigne par C \mathbb{C} C R \R R R ⊂ C \R\subset \mathbb{C} R ⊂ C
C = { z = a + i b / a , b ∈ R } \bf \mathbb{C}=\{z=a+ib / a,b\in\R\} C = { z = a + ib/a , b ∈ R }
Exemples
Les nombres 1 + i 1+i 1 + i − 3 + 5 i -3+5i − 3 + 5 i 2 i 2i 2 i 1 − 3 i 1-\sqrt{3}i 1 − 3 i
Dans chacun des exemples suivants, donner la partie réelle et la partie imaginaire de z z z
z = 2 + 3 i z=2+3i z = 2 + 3 i z = − 1 + 1 2 i z=-1+\dfrac{1}{2}i z = − 1 + 2 1 i z = − i z=-i z = − i z = 2 z=2 z = 2 z = 4 i − 1 3 z=4i-\dfrac{1}{3} z = 4 i − 3 1
Correction
z = 2 + 3 i z=2+3i z = 2 + 3 i
R e ( z ) = 2 \mathcal{R}e(z)=2 R e ( z ) = 2 I m ( z ) = 3 \mathcal{I}m(z)=3 I m ( z ) = 3
z = − 1 + 1 2 i z=-1+\dfrac{1}{2}i z = − 1 + 2 1 i
R e ( z ) = − 1 \mathcal{R}e(z)=-1 R e ( z ) = − 1 I m ( z ) = 1 2 \mathcal{I}m(z)=\dfrac12 I m ( z ) = 2 1
z = − i = 0 + ( − 1 ) i z=-i=0+(-1)i z = − i = 0 + ( − 1 ) i
R e ( z ) = 0 \mathcal{R}e(z)=0 R e ( z ) = 0 I m ( z ) = − 1 \mathcal{I}m(z)=-1 I m ( z ) = − 1
z = 2 = 2 + 0. i z=2=2+0.i z = 2 = 2 + 0. i
R e ( z ) = 2 \mathcal{R}e(z)=2 R e ( z ) = 2 I m ( z ) = 0 \mathcal{I}m(z)=0 I m ( z ) = 0
z = 4 i − 1 3 = − 1 3 + 4 i z=4i-\dfrac{1}{3}=-\dfrac13+4i z = 4 i − 3 1 = − 3 1 + 4 i
R e ( z ) = − 1 3 \mathcal{R}e(z)=-\dfrac13 R e ( z ) = − 3 1 I m ( z ) = 4 \mathcal{I}m(z)=4 I m ( z ) = 4
2) Aspet Géométrique
Le plan ( P ) \mathcal{(P)} ( P ) ( O , u ⃗ , v ⃗ ) (O,\vec{u},\vec{v}) ( O , u , v )
à tout nombre complexe z = x + y i z=x+yi z = x + y i M ( x , y ) M(x,y) M ( x , y )
que M M M z z z M ( z ) M(z) M ( z )
on dit que z z z M M M z M z_M z M z = a f f ( M ) z=aff(M) z = a ff ( M )
on dit que z z z O M → \overrightarrow{OM} OM z O M → = z z_{\overrightarrow{OM}}=z z OM = z
Le plan (P \mathcal{P} P
Le point O est l’image du nombre 0.
L’axe des abscisses ( O , u ⃗ ) (O,\vec{u}) ( O , u )
l’axe des ordonnées ( O , v ⃗ ) (O,\vec{v}) ( O , v )
Application
Place dans le plan complexe les points M i M_i M i z i z_i z i
z 1 = 2 + 3 i z_1=2+3i z 1 = 2 + 3 i z 2 = 3 + i z_2=3+i z 2 = 3 + i z 3 = − 1 + 2 i z_3=-1+2i z 3 = − 1 + 2 i z 4 = 2 − i z_4=2-i z 4 = 2 − i z 5 = i z_5=i z 5 = i z 6 = − 2 i z_6=-2i z 6 = − 2 i z 7 = − 2 z_7=-2 z 7 = − 2 z 8 = − i − 3 z_8=-i-3 z 8 = − i − 3
Correction
3) Propriétés
Proposition:
Deux nombres complexes sont égaux si et seulement si ils ont la même partie réelle et la même partie imaginaire :
z = z ′ ⟺ a + i b = a ′ + i b ′ ⟺ a = a ′ et b = b ′ . \begin{aligned}
z=z' \quad &\iff \quad a+ib=a'+ib' \\
\quad &\iff \quad a=a' \text{ et } b=b'.
\end{aligned} z = z ′ ⟺ a + ib = a ′ + i b ′ ⟺ a = a ′ et b = b ′ . Cas particulier : a + i b = 0 ⟺ a = 0 et b = 0 a+ib=0 \iff a=0 \text{ et } b=0 a + ib = 0 ⟺ a = 0 et b = 0
Proposition:
On pose z = a + i b z=a+ib z = a + ib z ′ = a ′ + i b ′ z'=a'+ib' z ′ = a ′ + i b ′ k ∈ R k\in\R k ∈ R
z + z ′ = ( a + a ′ ) + i ( b + b ′ ) z+z'=(a+a')+i(b+b') z + z ′ = ( a + a ′ ) + i ( b + b ′ ) z − z ′ = ( a − a ′ ) + i ( b − b ′ ) z-z'=(a-a')+i(b-b') z − z ′ = ( a − a ′ ) + i ( b − b ′ ) k z = k a + i k b kz=ka+ikb k z = ka + ikb z z ′ = ( a a ′ − b b ′ ) + i ( a b ′ + a ′ b ) zz'=(aa'-bb')+i(ab'+a'b) z z ′ = ( a a ′ − b b ′ ) + i ( a b ′ + a ′ b )
Preuve:
z . z ′ = ( a + i b ) ( a ′ + i b ′ ) = a a ′ + i a b ′ + i a ′ b + i 2 b b ′ = ( a a ′ − b b ′ ) + i ( a b ′ + a ′ b ) \begin{aligned}
z.z'&=(a+ib)(a'+ib')\\&=aa'+iab'+ia'b+i^2bb'\\&=(aa'-bb')+i(ab'+a'b)
\end{aligned} z . z ′ = ( a + ib ) ( a ′ + i b ′ ) = a a ′ + ia b ′ + i a ′ b + i 2 b b ′ = ( a a ′ − b b ′ ) + i ( a b ′ + a ′ b ) Proposition:
Soient z z z z ′ z' z ′
z z z I m ( z ) = 0 \mathcal{I}m(z)=0 I m ( z ) = 0 z z z R e ( z ) = 0 \mathcal{R}e(z)=0 R e ( z ) = 0 z = z ′ ⇔ { R e ( z ) = R e ( z ′ ) et I m ( z ) = I m ( z ′ ) z=z' \Leftrightarrow \left\{\begin{matrix}\mathcal{R}e(z)=\mathcal{R}e(z')\\ \text{ et } \mathcal{I}m(z)=\mathcal{I}m(z')\end{matrix}\right. z = z ′ ⇔ { R e ( z ) = R e ( z ′ ) et I m ( z ) = I m ( z ′ )
Exercice:
Ecrire sous forme algébrique les nombres complexes suivants :
a = ( 1 + 2 i ) 2 a=(1+2i)^2 a = ( 1 + 2 i ) 2 b = ( 5 + i ) ( 3 − 2 i ) b=(5+i)(3-2i) b = ( 5 + i ) ( 3 − 2 i ) c = 5 i ( 3 + 2 i ) c=5i(3+2i) c = 5 i ( 3 + 2 i )
Soit x x x z = x + 1 + i ( x 2 − x ) z=x+1+i(x^2-x) z = x + 1 + i ( x 2 − x )
a) Déterminer x pour que z soit réel
b) Déterminer x pour que z soit imaginaire pur.
Correction:
1.
a = ( 1 + 2 i ) 2 = 1 2 + 2 × 1 × 2 i + ( 2 i ) 2 = 1 + 4 i − 4 = − 3 + 4 i b = ( 5 + i ) ( 3 − 2 i ) = 15 − 10 i + 3 i − 2 i 2 = 15 − 7 i + 2 = 17 − 7 i c = 5 i ( 3 + 2 i ) = 15 i + 10 i 2 = − 10 + 15 i \begin{aligned}
a&=(1+2i)^2=1^2+2\times1\times2i+(2i)^2\\&=1+4i-4=-3+4i\\
b&=(5+i)(3-2i)\\&=15-10i+3i-2i^2\\&=15-7i+2=17-7i\\
c&=5i(3+2i)=15i+10i^2\\&=-10+15i
\end{aligned} a b c = ( 1 + 2 i ) 2 = 1 2 + 2 × 1 × 2 i + ( 2 i ) 2 = 1 + 4 i − 4 = − 3 + 4 i = ( 5 + i ) ( 3 − 2 i ) = 15 − 10 i + 3 i − 2 i 2 = 15 − 7 i + 2 = 17 − 7 i = 5 i ( 3 + 2 i ) = 15 i + 10 i 2 = − 10 + 15 i
Soit x ∈ R x\in\R x ∈ R z = x + 1 + i ( x 2 − x ) z=x+1+i(x^2-x) z = x + 1 + i ( x 2 − x )
a/
z ∈ R ⟺ I m ( z ) = 0 ⟺ x 2 − x = 0 ⟺ x ( x − 1 ) = 0 ⟺ x = 0 ou x = 1 \begin{aligned}
z\in\R &\Longleftrightarrow \mathcal{I}m(z)=0 \\
&\Longleftrightarrow x^2-x=0 \\
&\Longleftrightarrow x(x-1)=0 \\
&\Longleftrightarrow x=0 \text{ ou } x=1
\end{aligned} z ∈ R ⟺ I m ( z ) = 0 ⟺ x 2 − x = 0 ⟺ x ( x − 1 ) = 0 ⟺ x = 0 ou x = 1
b/
z ∈ i R ⟺ R e ( z ) = 0 ⟺ x + 1 = 0 ⟺ x = − 1 \begin{aligned}
z\in i\R &\Longleftrightarrow \mathcal{R}e(z)=0 \\
&\Longleftrightarrow x+1=0 \\
&\Longleftrightarrow x=-1
\end{aligned} z ∈ i R ⟺ R e ( z ) = 0 ⟺ x + 1 = 0 ⟺ x = − 1
Proposition:
Si A A A B B B C C C z A z_A z A z B z_B z B z C z_C z C
l’affixe du vecteur A B → \overrightarrow{AB} A B z B − z A z_B-z_A z B − z A A B = ∣ ∣ A B → ∣ ∣ = ∣ z B − z A ∣ AB=||\overrightarrow{AB}||=|z_B-z_A| A B = ∣∣ A B ∣∣ = ∣ z B − z A ∣
l’affixe de I I I [ A B ] [AB] [ A B ] z I = z A + z B 2 z_I=\dfrac{z_A+z_B}{2} z I = 2 z A + z B
Si A ≠ C A\ne C A = C A A A B B B C C C z B − z A z C − z A ∈ R \dfrac{z_B-z_A}{z_C-z_A}\in\R z C − z A z B − z A ∈ R
Preuve:
A B → = O B → − O A → ⟹ z A B → = z B − z A \overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA} \implies z_{\overrightarrow{AB}}=z_B-z_A A B = OB − O A ⟹ z A B = z B − z A I I I [ A B ] [AB] [ A B ]
⟹ I A → + I B → = 0 ⃗ ⟹ z A − z I + z B − z I = 0 ⟹ 2 z I = z A + z B \begin{aligned}
&\implies\overrightarrow{IA}+\overrightarrow{IB}=\vec{0}\\
&\implies z_A-z_I+z_B-z_I=0\\&
\implies 2z_I=z_A+z_B
\end{aligned} ⟹ I A + I B = 0 ⟹ z A − z I + z B − z I = 0 ⟹ 2 z I = z A + z B les points A,B et C sont align e ˊ s ⟺ ∃ k ∈ R : A B → = k A C → \begin{align*}
&\text{les points A,B et C sont alignés }\\
&\iff \exists k \in \R ~ : ~ \overrightarrow{AB}=k\overrightarrow{AC}
\end{align*} les points A,B et C sont align e ˊ s ⟺ ∃ k ∈ R : A B = k A C A B → = k A C → ⟺ z B − z A = k ( z C − z A ) ⟺ z B − z A z C − z A = k ⟺ z B − z A z C − z A ∈ R \begin{align*}
\overrightarrow{AB}=k\overrightarrow{AC}&\iff z_B-z_A=k(z_C-z_A)\\
&\iff \dfrac{z_B-z_A}{z_C-z_A}=k\\
&\iff \dfrac{z_B-z_A}{z_C-z_A}\in \R
\end{align*} A B = k A C ⟺ z B − z A = k ( z C − z A ) ⟺ z C − z A z B − z A = k ⟺ z C − z A z B − z A ∈ R Exercice: Soient A et B et C trois points dont les affixes respectivement sont :
z A = − 1 + i , z B = 2 i et z C = 2 − 2 i z_A=-1+i~~~,~~~ z_B=2i ~~~ \text{ et } ~~~ z_C=2-2i z A = − 1 + i , z B = 2 i et z C = 2 − 2 i
Déterminer l’affixe de I I I [ B C ] [BC] [ BC ]
Les points A A A B B B C C C
Calculer ∣ z A − z B ∣ |z_A-z_B| ∣ z A − z B ∣ ∣ z A − z C ∣ |z_A-z_C| ∣ z A − z C ∣ ∣ z C − z B ∣ |z_C-z_B| ∣ z C − z B ∣
Correction:
z I = z B + z C 2 = 2 i + 2 − 2 i 2 = 1 z_I=\dfrac{z_B+z_C}{2}=\dfrac{2i+2-2i}{2}=1 z I = 2 z B + z C = 2 2 i + 2 − 2 i = 1
z B − z A z C − z A = 2 i + 1 − i 2 − 2 i + 1 − i = i + 1 3 − 3 i = ( i + 1 ) ( 3 + 3 i ) ( 3 − 3 i ) ( 3 + 3 i ) = 3 i − 3 + 3 + 3 i 3 2 − ( 3 i ) 2 = 6 i 9 + 9 = 1 3 i \begin{align*}
\dfrac{z_B-z_A}{z_C-z_A}&=\dfrac{2i+1-i}{2-2i+1-i}\\
&=\dfrac{i+1}{3-3i}=\dfrac{(i+1)(3+3i)}{(3-3i)(3+3i)}
\\&=\dfrac{3i-3+3+3i}{3^2-(3i)^2}
\\&=\dfrac{6i}{9+9}=\dfrac13 i
\end{align*} z C − z A z B − z A = 2 − 2 i + 1 − i 2 i + 1 − i = 3 − 3 i i + 1 = ( 3 − 3 i ) ( 3 + 3 i ) ( i + 1 ) ( 3 + 3 i ) = 3 2 − ( 3 i ) 2 3 i − 3 + 3 + 3 i = 9 + 9 6 i = 3 1 i Comme z B − z A z C − z A ∉ R \dfrac{z_B-z_A}{z_C-z_A}\notin\R z C − z A z B − z A ∈ / R A A A B B B C C C
∣ z A − z B ∣ = ∣ − 1 + i − 2 i ∣ = ∣ − 1 − i ∣ = ( − 1 ) 2 + ( − 1 ) 2 = 2 \begin{align*}
|z_A-z_B|&=|-1+i-2i|\\&=|-1-i|\\&=\sqrt{(-1)^2+(-1)^2}\\&=\sqrt{2}\\
\end{align*} ∣ z A − z B ∣ = ∣ − 1 + i − 2 i ∣ = ∣ − 1 − i ∣ = ( − 1 ) 2 + ( − 1 ) 2 = 2 ∣ z A − z C ∣ = ∣ − 1 + i − 2 + 2 i ∣ = ∣ − 3 + 3 i ∣ = ( − 3 ) 2 + ( 3 ) 2 = 18 \begin{align*}
|z_A-z_C|&=|-1+i-2+2i|\\&=|-3+3i|\\&=\sqrt{(-3)^2+(3)^2}\\&=\sqrt{18}
\end{align*} ∣ z A − z C ∣ = ∣ − 1 + i − 2 + 2 i ∣ = ∣ − 3 + 3 i ∣ = ( − 3 ) 2 + ( 3 ) 2 = 18 ∣ z C − z B ∣ = ∣ 2 − 2 i − 2 i ∣ = ∣ 2 − 4 i ∣ = ( 2 ) 2 + ( − 4 ) 2 = 20 \begin{align*}
|z_C-z_B|&=|2-2i-2i|\\&=|2-4i|\\&=\sqrt{(2)^2+(-4)^2}\\&=\sqrt{20}
\end{align*} ∣ z C − z B ∣ = ∣2 − 2 i − 2 i ∣ = ∣2 − 4 i ∣ = ( 2 ) 2 + ( − 4 ) 2 = 20 Conclusion :
A B = ∣ z A − z B ∣ = 2 AB=|z_A-z_B|=\sqrt{2} A B = ∣ z A − z B ∣ = 2 A C = ∣ z A − z C ∣ = 18 AC=|z_A-z_C|=\sqrt{18} A C = ∣ z A − z C ∣ = 18 B C = ∣ z C − z B ∣ = 20 BC=|z_C-z_B|=\sqrt{20} BC = ∣ z C − z B ∣ = 20
Donc A B 2 = 2 AB^2=2 A B 2 = 2 A C 2 = 18 AC^2=18 A C 2 = 18 B C 2 = 20 BC^2=20 B C 2 = 20
Alors A B 2 + A C 2 = B C 2 AB^2+AC^2=BC^2 A B 2 + A C 2 = B C 2
D’aprés la réciproque de Pythagore le triangle A B C ABC A BC A A A
4) Conjugué d’un nombre complexe
Définition
Le conjugu e ˊ \bf \text{conjugué} conjugu e ˊ z = a + i b z = a + i b z = a + ib z ˉ = a − i b \bar{z} = a - i b z ˉ = a − ib
Proposition :
Soit z = x + i y z=x+iy z = x + i y x x x y y y
z ˉ ˉ = z \bar{\bar{z}}=z z ˉ ˉ = z z + z ˉ = 2 R e ( z ) z+\bar{z}=2\mathcal{R}e(z) z + z ˉ = 2 R e ( z ) z − z ˉ = 2 i I m ( z ) z-\bar{z}=2i\mathcal{I}m(z) z − z ˉ = 2 i I m ( z ) z . z ˉ = x 2 + y 2 z.\bar{z}=x^2+y^2 z . z ˉ = x 2 + y 2 z ∈ R ⇔ z = z ˉ z\in\R \Leftrightarrow z=\bar{z} z ∈ R ⇔ z = z ˉ z ∈ i R ⇔ z = − z ˉ z\in i\R \Leftrightarrow z=-\bar{z} z ∈ i R ⇔ z = − z ˉ
Preuve :
Soit z = x + i y z=x+iy z = x + i y z ˉ = x − i y \bar{z}=x-iy z ˉ = x − i y
∙ z ˉ ˉ = x − i y ‾ = x + i ( − y ) ‾ = x − i ( − y ) = x + i y = z \begin{aligned}
\quad\bullet \quad
\bar{\bar{z}}&=\overline{x-iy}=\overline{x+i(-y)}\\&=x-i(-y)=x+iy=z
\end{aligned} ∙ z ˉ ˉ = x − i y = x + i ( − y ) = x − i ( − y ) = x + i y = z
∙ z + z ˉ = x + i y + x − i y = 2 x = 2 R e ( z ) \begin{aligned}
\quad\bullet \quad
z+\bar{z}&=x+iy+x-iy\\&=2x=2\mathcal{R}e(z)
\end{aligned} ∙ z + z ˉ = x + i y + x − i y = 2 x = 2 R e ( z )
∙ z − z ˉ = x + i y − ( x − i y ) = x + i y − x + i y = 2 i y = 2 i I m ( z ) \begin{aligned}
\quad\bullet \quad
z-\bar{z}&=x+iy-(x-iy)\\&=x+iy-x+iy\\&=2iy=2i\mathcal{I}m(z)
\end{aligned} ∙ z − z ˉ = x + i y − ( x − i y ) = x + i y − x + i y = 2 i y = 2 i I m ( z )
∙ z . z ˉ = ( x + i y ) ( x − i y ) = x 2 − i x y + i x y − i 2 y 2 = x 2 + y 2 \begin{aligned}
\quad\bullet \quad
z.\bar{z}&=(x+iy)(x-iy)\\&=x^2-ixy+ixy-i^2y^2\\&=x^2+y^2
\end{aligned} ∙ z . z ˉ = ( x + i y ) ( x − i y ) = x 2 − i x y + i x y − i 2 y 2 = x 2 + y 2
∙ z = z ˉ ⇔ x + i y = x − i y ⇔ 2 i y = 0 ⇔ y = 0 ⇔ I m ( z ) = 0 ⇔ z ∈ R \begin{aligned}
\quad\bullet \quad
z=\bar{z} &\Leftrightarrow x+iy=x-iy \\&\Leftrightarrow 2iy=0
\\&\Leftrightarrow y=0 \\
&\Leftrightarrow \mathcal{I}m(z)=0 \\
&\Leftrightarrow z\in\R
\end{aligned} ∙ z = z ˉ ⇔ x + i y = x − i y ⇔ 2 i y = 0 ⇔ y = 0 ⇔ I m ( z ) = 0 ⇔ z ∈ R
∙ z = − z ˉ ⇔ x + i y = − x + i y ⇔ 2 x = 0 ⇔ x = 0 ⇔ R e ( z ) = 0 ⇔ z ∈ i R \begin{aligned}
\quad\bullet \quad
z=-\bar{z} &\Leftrightarrow x+iy=-x+iy \\
&\Leftrightarrow 2x=0\\
&\Leftrightarrow x=0 \\
&\Leftrightarrow \mathcal{R}e(z)=0\\
&\Leftrightarrow z\in i\R
\end{aligned} ∙ z = − z ˉ ⇔ x + i y = − x + i y ⇔ 2 x = 0 ⇔ x = 0 ⇔ R e ( z ) = 0 ⇔ z ∈ i R
Proposition :
Soient z z z z ′ z' z ′ n n n
z + z ′ ‾ = z ˉ + z ′ ˉ \overline{z+z'}=\bar{z}+\bar{z'} z + z ′ = z ˉ + z ′ ˉ z . z ′ ‾ = z ˉ . z ′ ˉ \overline{z.z'}=\bar{z}.\bar{z'} z . z ′ = z ˉ . z ′ ˉ z n ‾ = ( z ˉ ) n \overline{z^n}=(\bar{z})^n z n = ( z ˉ ) n ( z z ′ ) ‾ = z ˉ z ′ ˉ \overline{(\dfrac{z}{z'})}=\dfrac{\bar{z}}{\bar{z'}} ( z ′ z ) = z ′ ˉ z ˉ z ′ ≠ 0 z'\ne0 z ′ = 0
Exercice :
Ecrire sous forme algébrique les nombres complexes suivants :
a = 2 3 + 2 i ; ; b = 1 + i 5 − 3 i a=\dfrac{2}{3+2i} ~~~ ;; ~~~ b=\dfrac{1+i}{5-3i} a = 3 + 2 i 2 ;; b = 5 − 3 i 1 + i
Correction :
a = 2 3 + 2 i = 2 ( 3 − 2 i ) ( 3 + 2 i ) ( 3 − 2 i ) = 6 − 4 i 3 2 − ( 2 i ) 2 = 6 − 4 i 9 − ( − 4 ) = 6 − 4 i 13 = 6 13 − 4 13 i \begin{aligned}
a&=\dfrac{2}{3+2i}=\dfrac{2(3-2i)}{(3+2i)(3-2i)}\\&=\dfrac{6-4i}{3^2-(2i)^2}=\dfrac{6-4i}{9-(-4)}\\&=\dfrac{6-4i}{13}=\dfrac{6}{13}-\dfrac{4}{13}i
\end{aligned} a = 3 + 2 i 2 = ( 3 + 2 i ) ( 3 − 2 i ) 2 ( 3 − 2 i ) = 3 2 − ( 2 i ) 2 6 − 4 i = 9 − ( − 4 ) 6 − 4 i = 13 6 − 4 i = 13 6 − 13 4 i b = 1 + i 5 − 3 i = ( 1 + i ) ( 5 + 3 i ) ( 5 − 3 i ) ( 5 + 3 i ) = 5 + 3 i + 5 i + 3 i 2 5 2 − ( 3 i ) 2 = 5 + 8 i − 3 25 − ( − 9 ) = 2 + 8 i 34 = 2 34 + 8 34 i = 1 17 + 4 17 i \begin{aligned}
b&=\dfrac{1+i}{5-3i}=\dfrac{(1+i)(5+3i)}{(5-3i)(5+3i)}\\
&=\dfrac{5+3i+5i+3i^2}{5^2-(3i)^2}\\
&=\dfrac{5+8i-3}{25-(-9)}=\dfrac{2+8i}{34}\\
&=\dfrac{2}{34}+\dfrac{8}{34}i=\dfrac{1}{17}+\dfrac{4}{17}i
\end{aligned} b = 5 − 3 i 1 + i = ( 5 − 3 i ) ( 5 + 3 i ) ( 1 + i ) ( 5 + 3 i ) = 5 2 − ( 3 i ) 2 5 + 3 i + 5 i + 3 i 2 = 25 − ( − 9 ) 5 + 8 i − 3 = 34 2 + 8 i = 34 2 + 34 8 i = 17 1 + 17 4 i