Question

Qual a forma algébrica de:

A)(1-i/1+i)^18?
Resposta: -1

B) (-1+5i)^18/(2+3i)^18
Reposta: 512i

Obs: tentei multiplicar pelo conjugado mas acho que me perdi.

Obrigada

Answer 1

Resposta:

Letra A)  – 1

Letra B)  512i

Letra A)

Multiplicando a expressão (1 – i)/(1 + i) (base da potência) por (1 – i)/(1 – i) (quociente unitário, isto é, tem valor igual a um), onde z̅ = 1 – i é o complexo conjugado do denominador z = 1 + i, e recordando que i² = – 1, temos:

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\left[\left(\dfrac{1-i}{1+i}\right)\!\!\:\!\left(\dfrac{1-i}{1-i}\right)\right]^{\!18}$

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\left[\dfrac{(1-i)(1-i)}{(1+i)(1-i)\right]^{\!18}}$

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\left[\dfrac{(1-i)^2}{(1+i)(1-i)}\right]^{\!18}$

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\left(\dfrac{1-2i+i^2}{1^2-i^2}\right)^{\!\!18}$

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\left(\dfrac{1-2i+(-1)}{1-(-1)}\right)^{\!\!18}$

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\left(\dfrac{1-1-2i}{1+1}\right)^{\!\!18}$

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\left(\dfrac{-2\:\!i}{2}\right)^{\!\!18}$

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\:\!(-i)^{18}$

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\big[(-1)\cdot i\big]^{18}$

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\:\!(-1)^{18}\cdot i^{18}$

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\:\!1\cdot \!\:\!\left(i^2\right)^9$

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\:\!1\cdot (-1)^9$

$\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\:\!1\cdot (-1)$

$\boxed{\boldsymbol{\sf \left(\dfrac{1-i}{1+i}\right)^{\!\!18}=\:\!-1}}$

Letra B)

Reescrevendo a expressão (– 1 + 5i)¹⁸/(2 + 3i)¹⁸ e multiplicando (– 1 + 5i)/(2 + 3i) (base da potência que surgirá após a reescrita) por (2 – 3i)/(2 – 3i) (quociente unitário, ou seja, tem valor igual a um), onde z̅ = 2 – 3i é o complexo conjugado do denominador z = 2 + 3i, e relembrando que i² = – 1, obtemos:

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\left(\dfrac{-1+5i}{2+3i}\right)^{\!\!18}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\left(\dfrac{-1+5i}{2+3i}\:\!\cdot\:\!\dfrac{2-3i}{2-3i}\right)^{\!\!18}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\left[\dfrac{(-1+5i)(2-3i)}{(2+3i)(2-3i)}\right]^{\!18}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\left(\dfrac{(-1)\cdot 2+(-1)\cdot (-3i)+(5i)\cdot 2+5i\cdot (-3i)}{2^2-(3i)^2}\right)^{\!\!18}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\left(\dfrac{-2+3i+10i-15i^2}{4-3^2i^2}\right)^{\!\!18}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\left(\dfrac{-2+3i+10i-15(-1)}{4-9(-1)}\right)^{\!\!18}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\left(\dfrac{-2+15+13i}{4+9}\right)^{\!\!18}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\left(\dfrac{13+13i}{13}\right)^{\!\!18}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\left(\dfrac{13\:\!(i+i)}{13}\right)^{\!\!18}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\:\!(1+i)^{18}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\big[(1+i)^2\big]^{9}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\:\!(1^2+2i+i^2)^{9}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\big[1+2i+(-1)\big]^{9}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\:\!(2i)^{9}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\:\!2^9\cdot i^9$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\:\!512\cdot i^1\cdot i^8$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\:\!512\cdot i\cdot \left(i^2\right)^{\!4}$

$\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\:\!512\cdot i\cdot (-1)^{4}$

$\boxed{\boldsymbol{\sf \dfrac{(-1+5i)^{18}}{(2+3i)^{18}}\,=\:\!512i}}$