Página07
0175
Obtenha o conjugado de 1 + 3 i 2 − i \dfrac{1+3i}{2-i} 2 − i 1 + 3 i
0175 - Solução
Obtendo:
z = 1 + 3 i 2 − i × 2 + i 2 + i → … → z = − 1 5 + 7 5 i ⇒ z ‾ = − 1 5 − 7 5 i z=\dfrac{1+3i}{2-i}\times \dfrac{2+i}{2+i}\to\ldots\to z=-\dfrac{1}{5}+\dfrac{7}{5}i\Rightarrow \boxed{\overline{z}=-\dfrac{1}{5}-\dfrac{7}{5}i} z = 2 − i 1 + 3 i × 2 + i 2 + i → … → z = − 5 1 + 5 7 i ⇒ z = − 5 1 − 5 7 i
0174
Determinar o conjugado de 1 + i i \dfrac{1+i}{i} i 1 + i
0174 - Solução
Obtendo:
z = 1 + i i × − i − i → … → z = 1 − i ⇒ z ‾ = 1 + i z=\dfrac{1+i}{i}\times \dfrac{-i}{-i}\to\ldots\to z=1-i\Rightarrow \boxed{\overline{z}=1+i} z = i 1 + i × − i − i → … → z = 1 − i ⇒ z = 1 + i
0173
Calcule o conjugado do inverso do número complexo z = ( 1 + i 1 − i ) − 1 z=\left( \dfrac{1+i}{1-i} \right)^{-1} z = ( 1 − i 1 + i ) − 1
0173 - Solução
Calculando:
Se z = ( 1 + i 1 − i ) − 1 z=\left( \dfrac{1+i}{1-i} \right)^{-1} z = ( 1 − i 1 + i ) − 1 z = 1 − i 1 + i z=\dfrac{1-i}{1+i} z = 1 + i 1 − i
Se z = 1 − i 1 + i z=\dfrac{1-i}{1+i} z = 1 + i 1 − i 1 z = 1 + i 1 − i \dfrac{1}{z}=\dfrac{1+i}{1-i} z 1 = 1 − i 1 + i
Sendo 1 z = 1 + i 1 − i × 1 + i 1 + i → … → 1 z = i \dfrac{1}{z}=\dfrac{1+i}{1-i}\times\dfrac{1+i}{1+i}\to\ldots\to \dfrac{1}{z}=i z 1 = 1 − i 1 + i × 1 + i 1 + i → … → z 1 = i − i -i − i
0172
Sejam u u u v v v u 2 − v 2 = 6 u^2-v^2=6 u 2 − v 2 = 6 u ‾ + v ‾ = 1 − i \overline{u}+\overline{v}=1-i u + v = 1 − i u ‾ \overline{u} u v ‾ \overline{v} v u u u v v v u − v u-v u − v
0172 - Solução
Sendo u ∈ C u\in\mathbb{C} u ∈ C v ∈ C v\in\mathbb{C} v ∈ C u 2 − v 2 = 6 u^2-v^2=6 u 2 − v 2 = 6 u ‾ + v ‾ = 1 − i \overline{u}+\overline{v}=1-i u + v = 1 − i
adotando u = a + b i , a , b ∈ R u=a+bi,\,\,a,\,b\in\mathbb{R} u = a + bi , a , b ∈ R v = c + d i , c , d ∈ R v=c+di,\,\,c,\,d\in\mathbb{R} v = c + d i , c , d ∈ R
{ u + v = ( a + c ) + ( b + d ) i u ‾ + v ‾ = u + v ‾ = ( a + c ) − ( b + d ) i \left\{
\begin{array}{rcrcl}
u & + & v & = & (a+c)+(b+d)i \\
\overline{u} & + & \overline{v} & = & \overline{u+v}=(a+c)-(b+d)i
\end{array}\right. { u u + + v v = = ( a + c ) + ( b + d ) i u + v = ( a + c ) − ( b + d ) i
então:
{ a + c = 1 b + d = 1 \left\{\begin{array}{rcrcr}
a & + & c & = & 1\\
b & + & d & = & 1
\end{array}\right. { a b + + c d = = 1 1
u − v = ( u − v ) . u + v u + v = u 2 − v 2 u + v = 6 u + v = = 6 u + v × u + v ‾ u + v ‾ = 6 ( 1 − i ) ( 1 + i ) ( 1 − i ) … → u − v = 3 − 3 i \begin{array}{lcll}
u-v & = & (u-v).\dfrac{u+v}{u+v}=\dfrac{u^2-v^2}{u+v}=\dfrac{6}{u+v} & =\\
& & & \\
& = & \dfrac{6}{u+v}\times \dfrac{\overline{u+v}}{\overline{u+v}} =\dfrac{6(1-i)}{(1+i)(1-i)}\ldots & \to\boxed{u-v=3-3i}
\end{array} u − v = = ( u − v ) . u + v u + v = u + v u 2 − v 2 = u + v 6 u + v 6 × u + v u + v = ( 1 + i ) ( 1 − i ) 6 ( 1 − i ) … = → u − v = 3 − 3 i
0171
Dados os números complexos u = 1 + i u=1+i u = 1 + i v = 1 − i v=1-i v = 1 − i u 52 × v − 51 u^{52}\times v^{-51} u 52 × v − 51
0171 - Solução
Vamos calcular separadamente u 52 u^{52} u 52 v − 51 v^{-51} v − 51
u 52 = ( 1 + i ) 52 = [ ( 1 + i ) 2 ] 26 = ( 2 i ) 26 = 2 26 . ( i 4 ) 6 . i 2 − 1 → u 52 = − 2 26 u^{52}=(1+i)^{52}=[(1+i)^2]^{26}=(2i)^{26}=2^{26}.(i^4)^6.\cancel{i^2}^{\,-1}\to \boxed{u^{52}=-2^{26}} u 52 = ( 1 + i ) 52 = [( 1 + i ) 2 ] 26 = ( 2 i ) 26 = 2 26 . ( i 4 ) 6 . i 2 − 1 → u 52 = − 2 26
v − 51 = ( 1 − i ) − 51 = 1 51 [ ( 1 − i ) 2 ] 25 . i = 1 ( − 2 i ) 25 . i = − 1 2 25 . i 26 → v^{-51}=(1-i)^{-51}=\dfrac{1^{51}}{[(1-i)^2]^{25}.i}=\dfrac{1}{(-2i)^{25}.i}=-\dfrac{1}{2^{25}.i^{26}}\to v − 51 = ( 1 − i ) − 51 = [( 1 − i ) 2 ] 25 . i 1 51 = ( − 2 i ) 25 . i 1 = − 2 25 . i 26 1 →
v − 51 = − 1 2 25 . ( i 2 ) 13 → v − 51 = 2 − 25 v^{-51}=-\dfrac{1}{2^{25}.(i^2)^{13}}\to \boxed{v^{-51}=2^{-25}} v − 51 = − 2 25 . ( i 2 ) 13 1 → v − 51 = 2 − 25
u 52 × v − 51 = − 2 26 × 2 − 25 = − 2 u^{52}\times v^{-51}=-2^{26}\times 2^{-25}=\boxed{-2} u 52 × v − 51 = − 2 26 × 2 − 25 = − 2
0170
Variando o inteiro n n n ( 1 + i 1 − i ) n \left( \dfrac{1+i}{1-i}\right)^n ( 1 − i 1 + i ) n
0170 - Solução
Primeiramente, vamos operar a fração (executando a divisão); depois, vamos observar os valores obtidos, quando alterarmos o expoente inteiro n n n
1 + i 1 − i × 1 + i 1 + i = 2 i 2 = i \dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i}=\dfrac{2i}{2}=i 1 − i 1 + i × 1 + i 1 + i = 2 2 i = i i n i^n i n
Portanto, ( 1 + i 1 − i ) n = i n \left( \dfrac{1+i}{1-i}\right)^n=i^n ( 1 − i 1 + i ) n = i n i i i − i -i − i 1 1 1 − 1 -1 − 1
0169
Dê as condições necessárias e suficientes para que a + b i c + d i \quad\dfrac{a+bi}{c+di}\quad c + d i a + bi c + d i ≠ 0 c+di\neq0 c + d i = 0
a) imaginário puro;
b) real.
0169 - Solução
Primeiramente, vamos efetuar a divisão a + b i c + d i \dfrac{a+bi}{c+di} c + d i a + bi
a + b i c + d i × c − d i c − d i = a c − a d i + b c i + b d c 2 + d 2 = a c + b d c 2 + d 2 − a d − b c c 2 + d 2 i \dfrac{a+bi}{c+di}\times \dfrac{c-di}{c-di}=\dfrac{ac-adi+bci+bd}{c^2+d^2}=\dfrac{ac+bd}{c^2+d^2}-\dfrac{ad-bc}{c^2+d^2}i c + d i a + bi × c − d i c − d i = c 2 + d 2 a c − a d i + b c i + b d = c 2 + d 2 a c + b d − c 2 + d 2 a d − b c i c 2 + d 2 ≠ 0 c^2+d^2\neq 0 c 2 + d 2 = 0
a) Para ser imaginário puro, zeramos a parte real, ou seja: a c + b d = 0 → a c = − b d ac+bd=0\to \boxed{ac=-bd} a c + b d = 0 → a c = − b d
b) Para ser real, zeramos a parte imaginária, ou seja: a d − b c = 0 → a d = b c ad-bc=0\to \boxed{ad=bc} a d − b c = 0 → a d = b c
0168
Se u = x + i y u=x+iy u = x + i y v = 1 2 − i 3 2 v=\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2} v = 2 1 − i 2 3 v . u ‾ v.\overline{u} v . u
0168 - Solução
Calculando, teremos:
v . u ‾ = ( 1 2 − i 3 2 ) × ( x − i y ) = ( x 2 − y 3 2 ) − ( y 2 + x 3 2 ) i v.\overline{u}=\left(\dfrac{1}{2}-i\dfrac{\sqrt{3}}{2}\right)\times (x-iy)=\left( \dfrac{x}{2}-y\dfrac{\sqrt{3}}{2}\right)-\left( \dfrac{y}{2}+x\dfrac{\sqrt{3}}{2}\right)i v . u = ( 2 1 − i 2 3 ) × ( x − i y ) = ( 2 x − y 2 3 ) − ( 2 y + x 2 3 ) i
Então: R e ( z ) = ( x 2 − y 3 2 ) Re(z)=\left( \dfrac{x}{2}-y\dfrac{\sqrt{3}}{2}\right) R e ( z ) = ( 2 x − y 2 3 )
0167
Seja z = x + i y z=x+iy z = x + i y x 2 + y 2 ≠ 0 x^2+y^2\neq 0 x 2 + y 2 = 0 i 2 = − 1 , x e y reais i^2=-1,\,\,x\,\,\text{e}\,\,y\,\,\text{reais} i 2 = − 1 , x e y reais z + 1 z z+\dfrac{1}{z} z + z 1
0167 - Solução
Sendo z = x + i y z=x+iy z = x + i y x 2 + y 2 ≠ 0 x^2+y^2\neq 0 x 2 + y 2 = 0
u = z + 1 z = z 2 + 1 z = ( z 2 + 1 ) z ‾ z . z ‾ = ( x 2 + y 2 + 1 ) . x x 2 + y 2 + ( x 2 + y 2 − 1 ) . y x 2 + y 2 . i u=z+\dfrac{1}{z}=\dfrac{z^2+1}{z}=\dfrac{(z^2+1)\overline{z}}{z.\overline{z}}=\dfrac{(x^2+y^2+1).x}{x^2+y^2}+\dfrac{(x^2+y^2-1).y}{x^2+y^2}.i u = z + z 1 = z z 2 + 1 = z . z ( z 2 + 1 ) z = x 2 + y 2 ( x 2 + y 2 + 1 ) . x + x 2 + y 2 ( x 2 + y 2 − 1 ) . y . i
I m ( u ) = 0 → y = 0 Im(u)=0\to \boxed{y=0} I m ( u ) = 0 → y = 0 x 2 + y 2 = 1 \boxed{x^2+y^2=1} x 2 + y 2 = 1
0166
Se z 1 z_{1} z 1 z 2 z_{2} z 2 z 1 + z 2 z_{1}+z_{2} z 1 + z 2 z 1 . z 2 z_{1}.z_{2} z 1 . z 2 z 1 z_{1} z 1 z 2 z_{2} z 2
0166 - Solução
Sendo z 1 ∈ C z_{1}\in\mathbb{C} z 1 ∈ C z 2 ∈ C z_{2}\in\mathbb{C} z 2 ∈ C z 1 + z 2 z_{1}+z_{2} z 1 + z 2 z 1 . z 2 z_{1}.z_{2} z 1 . z 2
Adotando z 1 = a + b i z_{1}=a+bi z 1 = a + bi z 2 = c + d i z_{2}=c+di z 2 = c + d i
{ u = z 1 + z 2 = ( a + c ) + ( b + d ) i v = z 1 . z 2 = ( a c − b d ) + ( a d + b c ) i \left\{
\begin{array}{rcrcl}
u & = & z_{1}+z_{2} & = & (a+c)+(b+d)i \\
v & = & z_{1}.z_{2} & = & (ac-bd)+(ad+bc)i
\end{array}\right. { u v = = z 1 + z 2 z 1 . z 2 = = ( a + c ) + ( b + d ) i ( a c − b d ) + ( a d + b c ) i
Como z 1 + z 2 z_{1}+z_{2} z 1 + z 2 I m ( u ) = 0 → b = − d Im(u)=0\to b=-d I m ( u ) = 0 → b = − d
Como z 1 . z 2 z_{1}.z_{2} z 1 . z 2 I m ( v ) = 0 → a d = − b c Im(v)=0\to ad=-bc I m ( v ) = 0 → a d = − b c
e, resolvendo o sistema, teremos, analisando a primeira linha do sistema:
z 1 + z 2 = a + c z_{1}+z_{2}=a+c z 1 + z 2 = a + c
a) b = d = 0 → z 1 + z 2 = a + c b=d=0\to z_{1}+z_{2}=a+c b = d = 0 → z 1 + z 2 = a + c a ∈ R a\in\mathbb{R} a ∈ R c ∈ R c\in\mathbb{R} c ∈ R z 1 z_{1} z 1 z 2 z_{2} z 2
b) b = − d b=-d b = − d b ∈ R ∗ b\in\mathbb{R^{*}} b ∈ R ∗ d ∈ R ∗ d\in\mathbb{R^{*}} d ∈ R ∗ a = c a=c a = c z 1 z_{1} z 1 z 2 z_{2} z 2 z 1 = z ‾ 2 z_{1}=\overline{z}_{2} z 1 = z 2
0165
Determine x ( x ∈ R ) x\,(x\in\mathbb{R}) x ( x ∈ R ) z = 2 − x i 1 + 2 x i z=\dfrac{2-xi}{1+2xi} z = 1 + 2 x i 2 − x i
0165 - Solução
Determinando x ( x ∈ R ) x\,(x\in\mathbb{R}) x ( x ∈ R ) z = 2 − x i 1 + 2 x i z=\dfrac{2-xi}{1+2xi} z = 1 + 2 x i 2 − x i
z = 2 − x i 1 + 2 x i × 1 − 2 x i 1 − 2 x i = 2 − 2 x 2 1 + 4 x 2 − 5 x 1 + 4 x 2 . i z=\dfrac{2-xi}{1+2xi}\times \dfrac{1-2xi}{1-2xi}=\dfrac{2-2x^2}{1+4x^2}-\dfrac{5x}{1+4x^2}.i z = 1 + 2 x i 2 − x i × 1 − 2 x i 1 − 2 x i = 1 + 4 x 2 2 − 2 x 2 − 1 + 4 x 2 5 x . i
R e ( z ) = 0 → x = 1 Re(z)=0\to \boxed{x=1} R e ( z ) = 0 → x = 1 x = − 1 \boxed{x=-1} x = − 1
0164
Determine a ( a ∈ R ) a\,(a\in\mathbb{R}) a ( a ∈ R ) z = 1 + 2 i 2 + a i z=\dfrac{1+2i}{2+ai} z = 2 + ai 1 + 2 i
0164 - Solução
a ( a ∈ R ) a\,(a\in\mathbb{R}) a ( a ∈ R ) z = 1 + 2 i 2 + a i z=\dfrac{1+2i}{2+ai} z = 2 + ai 1 + 2 i
z = 1 + 2 i 2 + a i × 2 − a i 2 − a i = 2 + 2 a 4 + a 2 + 4 − a 4 + a 2 . i z=\dfrac{1+2i}{2+ai}\times \dfrac{2-ai}{2-ai}=\dfrac{2+2a}{4+a^2}+\dfrac{4-a}{4+a^2}.i z = 2 + ai 1 + 2 i × 2 − ai 2 − ai = 4 + a 2 2 + 2 a + 4 + a 2 4 − a . i
I m ( z ) = 0 → a = 4 Im(z)=0\to \boxed{a=4} I m ( z ) = 0 → a = 4
0163
Determine o número complexo cujo produto por 5 + 8 i 5+8i 5 + 8 i 1 + i 1+i 1 + i
0163 - Solução
Determinando o número complexo cujo produto por 5 + 8 i 5+8i 5 + 8 i 1 + i 1+i 1 + i
z = 5 + 8 i e z ′ = 1 + i z 1 = x + y i \left.
\begin{array}{lcl}
z & = & 5+8i\,\,\text{e}\,\,z'=1+i\\
z_{1} & = & x+yi\\
\end{array}\right. z z 1 = = 5 + 8 i e z ′ = 1 + i x + y i
I) u = z . z 1 = ( 5 x − 8 y ) + ( 8 x + 5 y ) i u=z.z_{1}=(5x-8y)+(8x+5y)i u = z . z 1 = ( 5 x − 8 y ) + ( 8 x + 5 y ) i
I m ( u ) = 0 → x = − 5 8 y Im(u)=0\to x=-\dfrac{5}{8}y I m ( u ) = 0 → x = − 8 5 y
II) v = z 1 z ′ = x + y i 1 + i × 1 − i 1 − i = … = x + y 2 + − x + y 2 i v=\dfrac{z_{1}}{z'}=\dfrac{x+yi}{1+i}\times \dfrac{1-i}{1-i}=\ldots=\dfrac{x+y}{2}+\dfrac{-x+y}{2}i v = z ′ z 1 = 1 + i x + y i × 1 − i 1 − i = … = 2 x + y + 2 − x + y i
R e ( v ) = 0 → x = − y Re(v)=0\to x=-y R e ( v ) = 0 → x = − y I m ( v ) ≠ 0 → x ≠ y Im(v)\neq 0\to x\neq y I m ( v ) = 0 → x = y
De (I) e (II) teremos: x = y = 0 x=y=0 x = y = 0
0162
Coloque na forma a + b i a+bi a + bi
a ) 1 i c ) 3 + 4 i 2 − i e ) i 11 + 2. i 13 i 18 − i 37 g ) i 3 − i 2 + i 17 − i 35 i 16 − i 13 + i 30 b ) 1 1 + i d ) 1 + i 1 − i f ) 1 − 3 i 3 − i h ) 1 + i ( 1 − i ) 2 \begin{array}{llllllll}
a) & \dfrac{1}{i} & c) & \dfrac{3+4i}{2-i} & e) & \dfrac{i^{11}+2.i^{13}}{i^{18}-i^{37}} & g) & \dfrac{i^3-i^2+i^{17}-i^{35}}{i^{16}-i^{13}+i^{30}}\\
& & & & & & &\\
b) & \dfrac{1}{1+i} & d) & \dfrac{1+i}{1-i} & f) & \dfrac{1-3i}{3-i} & h) & \dfrac{1+i}{(1-i)^2}
\end{array} a ) b ) i 1 1 + i 1 c ) d ) 2 − i 3 + 4 i 1 − i 1 + i e ) f ) i 18 − i 37 i 11 + 2. i 13 3 − i 1 − 3 i g ) h ) i 16 − i 13 + i 30 i 3 − i 2 + i 17 − i 35 ( 1 − i ) 2 1 + i
0162 - Solução
Colocando na forma a + b i a+bi a + bi
a) 1 i × − i − i = − i + 1 = − i \dfrac{1}{i}\times \dfrac{-i}{-i}=\dfrac{-i}{+1}=\boxed{-i} i 1 × − i − i = + 1 − i = − i
b) 1 1 + i × 1 − i 1 − i = 1 − i 2 = 1 2 − 1 2 i \dfrac{1}{1+i}\times \dfrac{1-i}{1-i}=\dfrac{1-i}{2}=\boxed{\dfrac{1}{2}-\dfrac{1}{2}i} 1 + i 1 × 1 − i 1 − i = 2 1 − i = 2 1 − 2 1 i
c) 3 + 4 i 2 − i × 2 + i 2 + i = 6 + 3 i + 8 i − 4 4 + 1 = 2 5 + 11 5 i \dfrac{3+4i}{2-i}\times \dfrac{2+i}{2+i}=\dfrac{6+3i+8i-4}{4+1}=\boxed{\dfrac{2}{5}+\dfrac{11}{5}i} 2 − i 3 + 4 i × 2 + i 2 + i = 4 + 1 6 + 3 i + 8 i − 4 = 5 2 + 5 11 i
d) 1 + i 1 − i × 1 + i 1 + i = 1 + i + i − 1 1 + 1 = i \dfrac{1+i}{1-i}\times \dfrac{1+i}{1+i}=\dfrac{1+i+i-1}{1+1}=\boxed{i} 1 − i 1 + i × 1 + i 1 + i = 1 + 1 1 + i + i − 1 = i
e) i 11 + 2. i 13 i 18 − i 37 = ( i 4 ) 2 . i 3 + 2 ( i 4 ) 3 . i ( i 4 ) 4 . i 2 − ( i 4 ) 9 . i = − i 1 + i × 1 − i 1 − i = … = − 1 2 − 1 2 i \dfrac{i^{11}+2.i^{13}}{i^{18}-i^{37}}=\dfrac{(i^4)^2.i^3+2(i^4)^3.i}{(i^4)^4.i^2-(i^4)^9.i}=\dfrac{-i}{1+i}\times \dfrac{1-i}{1-i}=\ldots=\boxed{-\dfrac{1}{2}-\dfrac{1}{2}i} i 18 − i 37 i 11 + 2. i 13 = ( i 4 ) 4 . i 2 − ( i 4 ) 9 . i ( i 4 ) 2 . i 3 + 2 ( i 4 ) 3 . i = 1 + i − i × 1 − i 1 − i = … = − 2 1 − 2 1 i
f) 1 − 3 i 3 − i × 3 + i 3 + 1 = 3 + i − 9 i + 3 9 + 1 = … = 3 5 − 4 5 i \dfrac{1-3i}{3-i}\times \dfrac{3+i}{3+1}=\dfrac{3+i-9i+3}{9+1}=\ldots=\boxed{\dfrac{3}{5}-\dfrac{4}{5}i} 3 − i 1 − 3 i × 3 + 1 3 + i = 9 + 1 3 + i − 9 i + 3 = … = 5 3 − 5 4 i
g) i 3 − i 2 + i 17 − i 35 i 16 − i 13 + i 30 = … = 1 + i − i × i i = … = − 1 + i \dfrac{i^3-i^2+i^{17}-i^{35}}{i^{16}-i^{13}+i^{30}}=\ldots=\dfrac{1+i}{-i}\times \dfrac{i}{i}=\ldots=\boxed{-1+i} i 16 − i 13 + i 30 i 3 − i 2 + i 17 − i 35 = … = − i 1 + i × i i = … = − 1 + i
h) 1 + i ( 1 − i ) 2 = … = 1 + i − 2 i × 2 i 2 i = … = − 1 2 + 1 2 i \dfrac{1+i}{(1-i)^2}=\ldots=\dfrac{1+i}{-2i}\times \dfrac{2i}{2i}=\ldots=\boxed{-\dfrac{1}{2}+\dfrac{1}{2}i} ( 1 − i ) 2 1 + i = … = − 2 i 1 + i × 2 i 2 i = … = − 2 1 + 2 1 i
0161
Coloque na forma algébrica os seguintes números:
a) 2 i \dfrac{2}{i}\quad i 2 3 2 + i \dfrac{3}{2+i}\quad 2 + i 3 1 + 2 i 3 − i \dfrac{1+2i}{3-i}\quad 3 − i 1 + 2 i i 9 4 − 3 i \dfrac{i^9}{4-3i} 4 − 3 i i 9
0161 - Soluções
Resolvendo:
a) 2 i × − i − i = − 2 i − i 2 = 2 i \dfrac{2}{i}\times \dfrac{-i}{-i}=\dfrac{-2i}{-i^2}=\boxed{2i} i 2 × − i − i = − i 2 − 2 i = 2 i
b) 3 2 + i × 2 − i 2 − i = 6 − 3 i 4 − i 2 = 6 5 − 3 5 i \dfrac{3}{2+i}\times \dfrac{2-i}{2-i}=\dfrac{6-3i}{4-i^2}=\boxed{\dfrac{6}{5}-\dfrac{3}{5}i} 2 + i 3 × 2 − i 2 − i = 4 − i 2 6 − 3 i = 5 6 − 5 3 i
c) 1 + 2 i 3 − i × 3 + i 3 + i = 1 + 7 i 9 − i 2 = 1 10 + 7 10 i \dfrac{1+2i}{3-i}\times \dfrac{3+i}{3+i}=\dfrac{1+7i}{9-i^2}=\boxed{\dfrac{1}{10}+\dfrac{7}{10}i} 3 − i 1 + 2 i × 3 + i 3 + i = 9 − i 2 1 + 7 i = 10 1 + 10 7 i
d) i 9 4 − 3 i × 4 + 3 i 4 + 3 i = 4 i − 3 16 − 9 i 2 = − 3 25 + 4 25 i \dfrac{i^9}{4-3i}\times \dfrac{4+3i}{4+3i}=\dfrac{4i-3}{16-9i^2}=\boxed{-\dfrac{3}{25}+\dfrac{4}{25}i} 4 − 3 i i 9 × 4 + 3 i 4 + 3 i = 16 − 9 i 2 4 i − 3 = − 25 3 + 25 4 i
0160
Qual a condição para que o número ( a + b i ) 4 (a+bi)^4 ( a + bi ) 4 a ∈ R a\in\mathbb{R} a ∈ R b ∈ R b\in\mathbb{R} b ∈ R estritamente negativo?
0160 - Solução
Resolvendo:
z = ( a + b i ) 4 = ( a 4 + b 4 − 6 a 2 b 2 ) + 4 a b ( a 2 − b 2 ) i z=(a+bi)^4=(a^4+b^4-6a^2b^2)+4ab(a^2-b^2)i z = ( a + bi ) 4 = ( a 4 + b 4 − 6 a 2 b 2 ) + 4 ab ( a 2 − b 2 ) i
R e ( z ) < 0 → a 4 + b 4 − 6 a 2 b 2 < 0 ( I ) Re(z)<0\to a^4+b^4-6a^2b^2<0\quad (I) R e ( z ) < 0 → a 4 + b 4 − 6 a 2 b 2 < 0 ( I )
I m ( z ) = 0 → 4 a b ( a 2 − b 2 ) = 0 ( I I ) Im(z)=0\to 4ab(a^2-b^2)=0\quad (II) I m ( z ) = 0 → 4 ab ( a 2 − b 2 ) = 0 ( II )
De ( I I ) (II) ( II ) a = 0 a=0 a = 0 b = 0 b=0 b = 0 a 2 − b 2 = 0 a^2-b^2=0 a 2 − b 2 = 0
Se a = 0 a=0 a = 0 ( I ) (I) ( I ) b 4 < 0 → b^4<0\to b 4 < 0 → impossível
Se b = 0 b=0 b = 0 ( I ) (I) ( I ) a 4 < 0 → a^4<0\to a 4 < 0 → impossível
Se a 2 − b 2 = 0 → a 2 = b 2 = k a^2-b^2=0\to a^2=b^2=k a 2 − b 2 = 0 → a 2 = b 2 = k k k k ( I ) (I) ( I )
k 2 + k 2 − 6 k 2 < 0 → − 4 k 2 < 0 k^2+k^2-6k^2<0\to -4k^2<0 k 2 + k 2 − 6 k 2 < 0 → − 4 k 2 < 0 k k k
Devemos ter a ≠ 0 a\neq0 a = 0 b ≠ 0 b\neq0\,\, b = 0 a 2 = b 2 \,\,a^2=b^2 a 2 = b 2 a b ≠ 0 \boxed{ab\neq0} ab = 0 a = ± b \boxed{a=\pm b} a = ± b
0159
Qual é a condição para que o produto de dois números complexos a + b i a+bi a + bi c + d i c+di c + d i
0159 - Solução
Efetuando: ( a + b i ) ( c + d i ) = ( a c − b d ) + ( a d + b c ⏟ = 0 ) i → a d + b c = 0 (a+bi)(c+di)=(ac-bd)+(\underbrace{ad+bc}_{=0})i\to \boxed{ad+bc=0} ( a + bi ) ( c + d i ) = ( a c − b d ) + ( = 0 a d + b c ) i → a d + b c = 0
0158
Quais os números complexos x x x y y y x + y i = i x+yi=i x + y i = i x i + y = 2 i − 1 xi+y=2i-1 x i + y = 2 i − 1
0158 - Resposta
x = 1 + i e y = i x=1+i\,\,\text{e}\,\,y=i x = 1 + i e y = i
0158 - Solução
Vamos adotar os números complexos x = a + b i \boxed{x=a+bi} x = a + bi y = c + d i \boxed{y=c+di} y = c + d i
a + b i + ( c + d i ) i = i → a + b i + c i − d = i → I = { a − d = 0 b + c = 1 a+bi+(c+di)i=i\to a+bi+ci-d=i\to I = \left\{
\begin{array}{rcrcrr}
a & - & d & = & 0 &\\
b & + & c & = & 1 &
\end{array}\right. a + bi + ( c + d i ) i = i → a + bi + c i − d = i → I = { a b − + d c = = 0 1
( a + b i ) i + c + d i = 2 i − 1 → a i − b + c + d i = − 1 + 2 i → I I = { − b + c = − 1 a + d = 2 (a+bi)i+c+di=2i-1\to ai-b+c+di=-1+2i\to II = \left\{
\begin{array}{rcrcrr}
-b & + & c & = & -1 &\\
a & + & d & = & 2 &
\end{array}
\right. ( a + bi ) i + c + d i = 2 i − 1 → ai − b + c + d i = − 1 + 2 i → II = { − b a + + c d = = − 1 2
De I I I I I II II I I I III III I V IV I V x x x y y y
I I I = { a + d = 2 a − d = 0 → a = 1 III=\left\{
\begin{array}{rcrcrr}
a & + & d & = & 2 &\\
a & - & d & = & 0 &
\end{array}\right.\rightarrow \boxed{a=1} III = { a a + − d d = = 2 0 → a = 1 d = 1 \boxed{d=1} d = 1
I V = { b + c = 1 − b + c = − 1 → b = 1 IV=\left\{
\begin{array}{rcrcrr}
b & + & c & = & 1 &\\
-b & + & c & = & -1 &
\end{array}\right.\rightarrow \boxed{b=1} I V = { b − b + + c c = = 1 − 1 → b = 1 c = 0 \boxed{c=0} c = 0
Portanto: x = 1 + i \boxed{x=1+i} x = 1 + i y = i \boxed{y=i} y = i
0157
Determine x ∈ R x\in\mathbb{R} x ∈ R y ∈ R y\in\mathbb{R} y ∈ R
a ) 3 + 5 i x = y − 15 i d ) ( x + y i ) 2 = 2 i b ) ( x + y i ) ( 2 + 3 i ) = 1 + 8 i e ) ( 2 − x + 3 y ) + 2 y i = 0 c ) ( 3 + y i ) + ( x − 2 i ) = 7 − 5 i f ) ( 3 − i ) ( x + y i ) = 20 \begin{array}{llll}
a) & 3+5ix=y-15i & d) & (x+yi)^2=2i\\
b) & (x+yi)(2+3i)=1+8i & e) & (2-x+3y)+2yi=0\\
c) & (3+yi)+(x-2i)=7-5i & f) & (3-i)(x+yi)=20
\end{array} a ) b ) c ) 3 + 5 i x = y − 15 i ( x + y i ) ( 2 + 3 i ) = 1 + 8 i ( 3 + y i ) + ( x − 2 i ) = 7 − 5 i d ) e ) f ) ( x + y i ) 2 = 2 i ( 2 − x + 3 y ) + 2 y i = 0 ( 3 − i ) ( x + y i ) = 20
0157 - Soluções
Aplicando a definição de igualdade, no campo dos números complexos, teremos:
α + β i = γ + δ i ⟺ α = γ e β = δ \boxed{\alpha+\beta i=\gamma+\delta i\quad \Longleftrightarrow \quad \alpha=\gamma\,\,\text{e}\,\,\beta=\delta} α + β i = γ + δ i ⟺ α = γ e β = δ a) 3 + 5 i x = y − 15 i ⟺ { 3 = y 5 x = − 15 ⟹ x = − 3 e y = 3 3+5ix=y-15i\Longleftrightarrow
\left\{\begin{array}{rcr}
3 & = & y \\
5x & = & -15
\end{array}\right.\Longrightarrow \boxed{x=-3}\,\,\text{e}\,\,\boxed{y=3} 3 + 5 i x = y − 15 i ⟺ { 3 5 x = = y − 15 ⟹ x = − 3 e y = 3
b) ( x + y i ) ( 2 + 3 i ) = 1 + 8 i → ( 2 x − 3 y ) + ( 3 x + 2 y ) i = 1 + 8 i → (x+yi)(2+3i)=1+8i\to (2x-3y)+(3x+2y)i=1+8i\to ( x + y i ) ( 2 + 3 i ) = 1 + 8 i → ( 2 x − 3 y ) + ( 3 x + 2 y ) i = 1 + 8 i →
{ 2 x − 3 y = 1 ( − 3 L 1 + 2 L 2 ) 3 x + 2 y = 8 \left\{
\begin{array}{rcrcrr}
2x & - & 3y & = & 1 & (-3L_{1}+2L_{2})\\
3x & + & 2y & = & 8 &
\end{array}\right. { 2 x 3 x − + 3 y 2 y = = 1 8 ( − 3 L 1 + 2 L 2 )
{ 2 x − 3 y = 1 13 y = 13 → y = 1 \left\{
\begin{array}{rcrcrr}
2x & - & 3y & = & 1 & \\
& & 13y & = & 13 & \rightarrow \boxed{y=1}
\end{array}\right. { 2 x − 3 y 13 y = = 1 13 → y = 1
Aplicando y = 1 y=1 y = 1
2 x − 3.1 = 1 → 2 x = 4 → x = 2 2x-3.1=1\to 2x=4\to \boxed{x=2} 2 x − 3.1 = 1 → 2 x = 4 → x = 2
c) ( 3 + y i ) + ( x − 2 i ) = 7 − 5 i → ( 3 + x ) + ( y − 2 ) i = 7 − 5 i (3+yi)+(x-2i)=7-5i\to (3+x)+(y-2)i=7-5i ( 3 + y i ) + ( x − 2 i ) = 7 − 5 i → ( 3 + x ) + ( y − 2 ) i = 7 − 5 i
{ x + 3 = 7 → x = 4 y − 2 = − 5 → y = − 3 \left\{
\begin{array}{rcrcrr}
x & + & 3 & = & 7 & \rightarrow \boxed{x=4} \\
& & & & & \\
y & - & 2 & = & -5 & \rightarrow \boxed{y=-3}
\end{array}\right. ⎩ ⎨ ⎧ x y + − 3 2 = = 7 − 5 → x = 4 → y = − 3
d) ( x 2 − y 2 ) + 2 x y i = 2 i ⟺ { x 2 − y 2 = 0 2 x y = 2 (x^2-y^2)+2xyi=2i\Longleftrightarrow
\left\{\begin{array}{rcr}
x^2-y^2 & = & 0 \\
2xy & = & 2
\end{array}\right. ( x 2 − y 2 ) + 2 x y i = 2 i ⟺ { x 2 − y 2 2 x y = = 0 2
Da primeira equação, vamos obter x = ± y x=\pm y x = ± y
2 ( ± y ) = 2 → ± 2 y 2 = 2 → y = ± 1 2(\pm y)=2\to \pm2y^2=2\to y=\pm\sqrt{1} 2 ( ± y ) = 2 → ± 2 y 2 = 2 → y = ± 1 x = ± 1 x=\pm\sqrt{1} x = ± 1
Ao final, teremos como solução, dois pares ordenados( x ; y ) (x;\,y) ( x ; y ) ( 1 ; 1 ) (1;\,1) ( 1 ; 1 ) ( − 1 ; − 1 ) (-1;\,-1) ( − 1 ; − 1 )
e) ( 2 − x + 3 y ) + 2 y i = 0 + 0 i (2-x+3y)+2yi=0+0i ( 2 − x + 3 y ) + 2 y i = 0 + 0 i
{ − x + 3 y = − 2 2 y = 0 → y = 0 \left\{
\begin{array}{rcrcrr}
-x & + & 3y & = & -2 & \\
& & 2y & = & 0 & \rightarrow \boxed{y=0}
\end{array}\right. { − x + 3 y 2 y = = − 2 0 → y = 0
Aplicando y = 0 y=0 y = 0
− x + 3.0 = − 2 → − x = − 2 → x = 2 -x+\cancel{3.0}=-2\to -x=-2\to \boxed{x=2} − x + 3.0 = − 2 → − x = − 2 → x = 2
f) ( 3 − i ) ( x + y i ) = 20 → ( 3 x + y ) + ( − x + 3 y ) i = 20 + 0 i (3-i)(x+yi)=20\to (3x+y)+(-x+3y)i=20+0i ( 3 − i ) ( x + y i ) = 20 → ( 3 x + y ) + ( − x + 3 y ) i = 20 + 0 i
{ 3 x + y = 20 − x + 3 y = 0 → x = 3 y \left\{
\begin{array}{rcrcrr}
3x & + & y & = & 20 &\\
-x & + & 3y & = & 0 & \rightarrow \boxed{x=3y}
\end{array}\right. { 3 x − x + + y 3 y = = 20 0 → x = 3 y
Aplicando x = 3 y x=3y x = 3 y
3.3 y + y = 20 → 10 y = 20 → y = 2 3.3y+y=20\to 10y=20\to \boxed{y=2} 3.3 y + y = 20 → 10 y = 20 → y = 2 x = 6 \boxed{x=6} x = 6
0156
Determine x ∈ R x\in\mathbb{R} x ∈ R y ∈ R y\in\mathbb{R} y ∈ R
a ) 2 + 3 y i = x + 9 i b ) ( x + y i ) ( 3 + 4 i ) = 7 + 26 i c ) ( x + y i ) 2 = 4 i \begin{array}{ll}
a) & 2+3yi=x+9i\\
b) & (x+yi)(3+4i)=7+26i\\
c) & (x+yi)^2=4i
\end{array} a ) b ) c ) 2 + 3 y i = x + 9 i ( x + y i ) ( 3 + 4 i ) = 7 + 26 i ( x + y i ) 2 = 4 i
0156 - Respostas
a) x = 2 e y = 3 x=2\,\,\text{e}\,\,y=3 x = 2 e y = 3
b) x = 5 e y = 2 x=5\,\,\text{e}\,\,y=2 x = 5 e y = 2
c) Dois pares ordenados( x ; y ) (x;\,y) ( x ; y ) ( 2 ; 2 ) (\sqrt{2};\,\sqrt{2}) ( 2 ; 2 ) ( − 2 ; − 2 ) (-\sqrt{2};\,-\sqrt{2}) ( − 2 ; − 2 )
0156 - Soluções
Aplicando a definição de igualdade, no campo dos números complexos:
α + β i = γ + δ i ⟺ α = γ e β = δ \boxed{\alpha+\beta i=\gamma+\delta i\quad \Longleftrightarrow \quad \alpha=\gamma\,\,\text{e}\,\,\beta=\delta} α + β i = γ + δ i ⟺ α = γ e β = δ a) 2 + 3 y i = x + 9 i ⟺ { 2 = x 3 y = 9 ⟹ x = 2 e y = 3 2+3yi=x+9i\Longleftrightarrow
\left\{\begin{array}{rcr}
2 & = & x \\
3y & = & 9
\end{array}\right.\Longrightarrow \boxed{x=2}\,\,\text{e}\,\,\boxed{y=3} 2 + 3 y i = x + 9 i ⟺ { 2 3 y = = x 9 ⟹ x = 2 e y = 3
b) ( x + y i ) ( 3 + 4 i ) = 7 + 26 i → ( 3 x − 4 y ) + ( 4 x + 3 y ) i = 7 + 26 i → (x+yi)(3+4i)=7+26i\to (3x-4y)+(4x+3y)i=7+26i\to ( x + y i ) ( 3 + 4 i ) = 7 + 26 i → ( 3 x − 4 y ) + ( 4 x + 3 y ) i = 7 + 26 i →
{ 3 x − 4 y = 7 ( − 4 L 1 + 3 L 2 ) 4 x + 3 y = 26 \left\{\begin{array}{rcrcrr}
3x & - & 4y & = & 7 & (-4L_{1}+3L_{2})\\
4x & + & 3y & = & 26 &
\end{array}\right. { 3 x 4 x − + 4 y 3 y = = 7 26 ( − 4 L 1 + 3 L 2 )
{ 3 x − 4 y = 7 25 y = 50 → y = 2 \left\{\begin{array}{rcrcrr}
3x & - & 4y & = & 7 & \\
& & 25y & = & 50 & \rightarrow \boxed{y=2}
\end{array}\right. { 3 x − 4 y 25 y = = 7 50 → y = 2
Aplicando y = 2 y=2 y = 2
3 x − 4.2 = 7 → 3 x = 7 + 8 → x = 5 3x-4.2=7\to 3x=7+8\to \boxed{x=5} 3 x − 4.2 = 7 → 3 x = 7 + 8 → x = 5
c) ( x 2 − y 2 ) + 2 x y i = 4 i ⟺ { x 2 − y 2 = 0 2 x y = 4 (x^2-y^2)+2xyi=4i\Longleftrightarrow
\left\{\begin{array}{rcr}
x^2-y^2 & = & 0 \\
2xy & = & 4
\end{array}\right. ( x 2 − y 2 ) + 2 x y i = 4 i ⟺ { x 2 − y 2 2 x y = = 0 4
Da primeira equação, vamos obter x = ± y x=\pm y x = ± y
2 ( ± y ) = 4 → ± 2 y 2 = 4 → y = ± 2 2(\pm y)=4\to \pm2y^2=4\to y=\pm\sqrt{2} 2 ( ± y ) = 4 → ± 2 y 2 = 4 → y = ± 2 x = ± 2 x=\pm\sqrt{2} x = ± 2
Ao final, teremos como solução, dois pares ordenados( x ; y ) (x;\,y) ( x ; y ) ( 2 ; 2 ) (\sqrt{2};\,\sqrt{2}) ( 2 ; 2 ) ( − 2 ; − 2 ) (-\sqrt{2};\,-\sqrt{2}) ( − 2 ; − 2 )
0155
A igualdade ( 1 + i ) n = ( 1 − i ) n (1+i)^n=(1-i)^n ( 1 + i ) n = ( 1 − i ) n
0155 - Resposta
Verifica-se quando "n n n 4 4 4
0155 - Solução
Observando algumas potências:
( 1 + i ) = 2 i (1+i)=2i ( 1 + i ) = 2 i
( 1 − i ) = − 2 i (1-i)=-2i ( 1 − i ) = − 2 i
( 1 + i ) 4 = ( 2 i ) 2 = − 4 (1+i)^4=(2i)^2=-4 ( 1 + i ) 4 = ( 2 i ) 2 = − 4
( 1 − i ) 4 = ( − 2 i ) 2 = − 4 (1-i)^4=(-2i)^2=-4 ( 1 − i ) 4 = ( − 2 i ) 2 = − 4
Portanto, ( 1 + i ) 4 = ( 1 − i ) 4 (1+i)^4=(1-i)^4 ( 1 + i ) 4 = ( 1 − i ) 4
I) Se "n n n 4 4 4 n = 4 p n=4p n = 4 p p p p
( 1 + i ) n = ( 1 + i ) 4 p = [ ( 1 + i ) 4 ] p = ( − 4 ) p (1+i)^n=(1+i)^{4p}=[(1+i)^4]^p=(-4)^p ( 1 + i ) n = ( 1 + i ) 4 p = [( 1 + i ) 4 ] p = ( − 4 ) p
( 1 − i ) n = ( 1 − i ) 4 p = [ ( 1 − i ) 4 ] p = ( − 4 ) p (1-i)^n=(1-i)^{4p}=[(1-i)^4]^p=(-4)^p ( 1 − i ) n = ( 1 − i ) 4 p = [( 1 − i ) 4 ] p = ( − 4 ) p
e daí ( 1 + i ) n = ( 1 − i ) n (1+i)^n=(1-i)^n ( 1 + i ) n = ( 1 − i ) n
II) Se "n n n 4 4 4 p p p n n n 4 4 4 1 1 1 2 2 2 3 3 3
1 º ) n = 4 p + 1 ( 1 + i ) n = ( 1 + i ) 4 p + 1 = ( 1 + i ) 4 p . ( 1 + i ) = ( 1 + i ) ( − 4 ) p ( 1 − i ) n = ( 1 − i ) 4 p + 1 = ( 1 − i ) 4 p . ( 1 − i ) = ( 1 − i ) ( − 4 ) p 2 º ) n = 4 p + 2 ( 1 + i ) n = ( 1 + i ) 4 p + 2 = ( 1 + i ) 4 p . ( 1 + i ) 2 = 2 i ( − 4 ) p ( 1 − i ) n = ( 1 − i ) 4 p + 2 = ( 1 − i ) 4 p . ( 1 − i ) 2 = − 2 i ( − 4 ) p 3 º ) n = 4 p + 3 ( 1 + i ) n = ( 1 + i ) 4 p + 3 = ( 1 + i ) 4 p . ( 1 + i ) 3 = ( − 2 + 2 i ) ( − 4 ) p ( 1 − i ) n = ( 1 − i ) 4 p + 3 = ( 1 − i ) 4 p . ( 1 − i ) 3 = ( − 2 − 2 i ) ( − 4 ) p \begin{array}{ll}
1º) & n=4p+1\\
& (1+i)^n=(1+i)^{4p+1}=(1+i)^{4p}.(1+i)=(1+i)(-4)^p\\
& (1-i)^n=(1-i)^{4p+1}=(1-i)^{4p}.(1-i)=(1-i)(-4)^p\\
&\\
2º) & n=4p+2\\
& (1+i)^n=(1+i)^{4p+2}=(1+i)^{4p}.(1+i)^2=2i(-4)^p\\
& (1-i)^n=(1-i)^{4p+2}=(1-i)^{4p}.(1-i)^2=-2i(-4)^p\\
&\\
3º) & n=4p+3\\
& (1+i)^n=(1+i)^{4p+3}=(1+i)^{4p}.(1+i)^3=(-2+2i)(-4)^p\\
& (1-i)^n=(1-i)^{4p+3}=(1-i)^{4p}.(1-i)^3=(-2-2i)(-4)^p
\end{array} 1º ) 2º ) 3º ) n = 4 p + 1 ( 1 + i ) n = ( 1 + i ) 4 p + 1 = ( 1 + i ) 4 p . ( 1 + i ) = ( 1 + i ) ( − 4 ) p ( 1 − i ) n = ( 1 − i ) 4 p + 1 = ( 1 − i ) 4 p . ( 1 − i ) = ( 1 − i ) ( − 4 ) p n = 4 p + 2 ( 1 + i ) n = ( 1 + i ) 4 p + 2 = ( 1 + i ) 4 p . ( 1 + i ) 2 = 2 i ( − 4 ) p ( 1 − i ) n = ( 1 − i ) 4 p + 2 = ( 1 − i ) 4 p . ( 1 − i ) 2 = − 2 i ( − 4 ) p n = 4 p + 3 ( 1 + i ) n = ( 1 + i ) 4 p + 3 = ( 1 + i ) 4 p . ( 1 + i ) 3 = ( − 2 + 2 i ) ( − 4 ) p ( 1 − i ) n = ( 1 − i ) 4 p + 3 = ( 1 − i ) 4 p . ( 1 − i ) 3 = ( − 2 − 2 i ) ( − 4 ) p
Concluímos que a igualdade ( 1 + i ) n = ( 1 − i ) n (1+i)^n=(1-i)^n ( 1 + i ) n = ( 1 − i ) n n n n 4 4 4
0154
Qual o resultado da simplificação de:
( 2 + i ) 101 . ( 2 − i ) 50 ( − 2 − i ) 100 . ( i − 2 ) 49 \dfrac{(2+i)^{101}.(2-i)^{50}}{(-2-i)^{100}.(i-2)^{49}} ( − 2 − i ) 100 . ( i − 2 ) 49 ( 2 + i ) 101 . ( 2 − i ) 50
0154 - Resposta
-5
0154 - Solução
Resolvendo, teremos:
( 2 + i ) . ( 2 + i ) 100 . ( 2 − i ) . ( 2 − i ) 49 ( − 1 ) 100 . ( 2 + i ) 100 . ( − 1 ) 49 . ( 2 − i ) 49 = ( 2 + i ) . ( 2 − i ) 1. ( − 1 ) = 4 + 1 − 1 = − 5 \dfrac{(2+i).\cancel{(2+i)^{100}}.(2-i).\cancel{(2-i)^{49}}}{(-1)^{100}.\cancel{(2+i)^{100}}.(-1)^{49}.\cancel{(2-i)^{49}}}=\dfrac{(2+i).(2-i)}{1.(-1)}=\dfrac{4+1}{-1}=\boxed{-5} ( − 1 ) 100 . ( 2 + i ) 100 . ( − 1 ) 49 . ( 2 − i ) 49 ( 2 + i ) . ( 2 + i ) 100 . ( 2 − i ) . ( 2 − i ) 49 = 1. ( − 1 ) ( 2 + i ) . ( 2 − i ) = − 1 4 + 1 = − 5
0153
Se i 2 = − 1 i^2=-1 i 2 = − 1 ( 1 + i ) 12 − ( 1 − i ) 12 (1+i)^{12}-(1-i)^{12} ( 1 + i ) 12 − ( 1 − i ) 12
0153 - Soluções
Calculando o valor:
( 1 + i ) 12 − ( 1 − i ) 12 = [ ( 1 + i ) 2 ] 6 − [ ( 1 − i ) 2 ] 6 = ( 2 i ) 6 − ( − 2 i ) 6 = 0 (1+i)^{12}-(1-i)^{12}=[(1+i)^2]^6-[(1-i)^2]^6=(2i)^6-(-2i)^6=0 ( 1 + i ) 12 − ( 1 − i ) 12 = [( 1 + i ) 2 ] 6 − [( 1 − i ) 2 ] 6 = ( 2 i ) 6 − ( − 2 i ) 6 = 0
0152
Prove que ( 1 − i ) 2 = − 2 i (1-i)^2=-2i ( 1 − i ) 2 = − 2 i ( 1 − i ) 96 + ( 1 − i ) 97 (1-i)^{96} + (1-i)^{97} ( 1 − i ) 96 + ( 1 − i ) 97
0152 - Soluções
Vamos dividir em duas questões:
a) Desenvolvendo o produto notável:
( 1 − i ) 2 = − 2 i → 1 2 − 2.1. i + i 2 = 1 − 1 − 2 i = − 2 i (1-i)^2=-2i\to 1^2-2.1.i+i^2=\cancel{1-1}-2i=-2i ( 1 − i ) 2 = − 2 i → 1 2 − 2.1. i + i 2 = 1 − 1 − 2 i = − 2 i
b) Calculando:
( 1 − i ) 96 + ( 1 − i ) 97 = [ ( 1 − i ) 2 ] 48 + [ ( 1 − i ) 2 ] 48 . ( 1 − i ) = (1-i)^{96} + (1-i)^{97}=[(1-i)^2]^{48}+[(1-i)^2]^{48}.(1-i)= ( 1 − i ) 96 + ( 1 − i ) 97 = [( 1 − i ) 2 ] 48 + [( 1 − i ) 2 ] 48 . ( 1 − i ) =
( − 2 i ) 48 + ( − 2 i ) 48 . ( 1 − i ) = ( − 2 i ) 48 [ 1 + 1. ( 1 − i ) ] = (-2i)^{48}+(-2i)^{48}.(1-i)=(-2i)^{48}[1+1.(1-i)]= ( − 2 i ) 48 + ( − 2 i ) 48 . ( 1 − i ) = ( − 2 i ) 48 [ 1 + 1. ( 1 − i )] =
( − 1 ) 48 .2 48 . [ ( i ) 4 ] 12 . ( 2 − i ) = 2 49 − i .2 48 (-1)^{48}.2^{48}.[(i)^4]^{12}.(2-i)=\boxed{2^{49}-i.2^{48}} ( − 1 ) 48 . 2 48 . [( i ) 4 ] 12 . ( 2 − i ) = 2 49 − i . 2 48
0151
Calcular as potências de "i i i
a ) i 76 b ) i 110 c ) i 97 d ) i 503 \begin{array}{llll}
a) i^{76} & b) i^{110} & c) i^{97} & d) i^{503}
\end{array} a ) i 76 b ) i 110 c ) i 97 d ) i 503
0151 - Soluções
Calculando as potências de "i i i
a) i 76 = ( i 4 ) 19 = 1 19 = 1 i^{76}=(i^{4})^{19}=1^{19}=1 i 76 = ( i 4 ) 19 = 1 19 = 1
b) i 110 = ( i 2 ) 55 = ( − 1 ) 55 = − 1 i^{110}=(i^{2})^{55}=(-1)^{55}=-1 i 110 = ( i 2 ) 55 = ( − 1 ) 55 = − 1
c) i 97 = i . i 96 = i . ( i 4 ) 24 = i .1 24 = i i^{97}=i.i^{96}=i.(i^{4})^{24}=i.1^{24}=i i 97 = i . i 96 = i . ( i 4 ) 24 = i . 1 24 = i
d) i 503 = i 3 . i 500 = − i . ( i 4 ) 125 = − i .1 125 = − i i^{503}=i^3.i^{500}=-i.(i^4)^{125}=-i.1^{125}=-i i 503 = i 3 . i 500 = − i . ( i 4 ) 125 = − i . 1 125 = − i
