Question

A) calcule os seguintes quocientes :​

Answer 1

$\Large\boxed{\begin{array}{l}\tt a)~\sf\dfrac{(2+3i)}{(2+i)}\cdot\dfrac{(2-i)}{(2-i)}=\dfrac{4-2i+6i-3i^2}{2^2-i^2}\\\\\sf\dfrac{4-2i+6i+3}{4-(-1)}=\dfrac{7+4i}{5}=\dfrac{7}{5}+\dfrac{4}{5}i\end{array}}$

$\Large\boxed{\begin{array}{l}\tt b)~\sf\dfrac{(4+2i)}{(2+i)}\cdot\dfrac{(2-i)}{(2-i)}=\dfrac{8-\diagdown\!\!\!\!4i+\diagdown\!\!\!\!4i-2i^2}{2^2-i^2}\\\\\sf\dfrac{8+2}{4+1}=\dfrac{10}{5}=2\end{array}}$

$\Large\boxed{\begin{array}{l}\tt c)~\sf\dfrac{(3-i)}{(1+i)}\cdot\dfrac{(1-i)}{(1-i)}=\dfrac{3-3i-i+i^2}{1^2-i^2}\\\\\sf\dfrac{3-3i-i-1}{1+1}=\dfrac{2-4i}{2}=1-2i\end{array}}$

$\Large\boxed{\begin{array}{l}\tt d)~\sf\dfrac{(1+i)}{(1-i)}\cdot\dfrac{(1+i)}{(1+i)}=\dfrac{1+2i+i^2}{1^2-i^2}\\\\\sf\dfrac{\backslash\!\!\!1+2i-\backslash\!\!\!1}{1+1}=\dfrac{2i}{2}=i\end{array}}$

$\Large\boxed{\begin{array}{l}\tt e)~\sf\dfrac{(2-5i)}{(2-7i)}\cdot\dfrac{(2+7i)}{(2+7i)}=\dfrac{4+14i-10i-35i^2}{2^2-(7i)^2}\\\\\sf\dfrac{4+14i-10i+35}{4+49}=\dfrac{39+4i}{53}=\dfrac{39}{53}+\dfrac{4}{53}i\end{array}}$

$\Large\boxed{\begin{array}{l}\tt f)~\sf\dfrac{(2-i)}{(1+i)}\cdot\dfrac{(1-i)}{(1-i)}=\dfrac{2-2i-i+i^2}{1^2-i^2}\\\\\sf\dfrac{2-2i-i-1}{1+1}=\dfrac{1-3i}{2}=\dfrac{1}{2}-\dfrac{3}{2}i\end{array}}$

$\Large\boxed{\begin{array}{l}\tt g)~\sf\dfrac{(1+i)}{(3-i)}\cdot\dfrac{(3+i)}{(3+i)}=\dfrac{3+i+3i+i^2}{3^2-i^2}\\\\\sf\dfrac{3+i+3i-1}{9+1}=\dfrac{2+4i}{10}=\dfrac{2}{10}+\dfrac{4}{10}i\\\\\sf\implies\dfrac{1}{5}+\dfrac{2}{5}i\end{array}}$

$\Large\boxed{\begin{array}{l}\tt h)~\sf\dfrac{(1+i)}{i}\cdot\dfrac{i}{i}=\dfrac{i+i^2}{i^2}=\dfrac{i-1}{-1}\\\\\sf\implies1-i\end{array}}$

$\large\boxed{\begin{array}{l}\tt j)~\sf\dfrac{(-4\sqrt{3}+4i)}{(\sqrt{3}+i)}\cdot\dfrac{(\sqrt{3}-i)}{(\sqrt{3}-i)}=\dfrac{-12+4\sqrt{3}i+4\sqrt{3}i-4i^2}{(\sqrt{3})^2+i^2}\\\\\sf\dfrac{-12+4\sqrt{3}+4\sqrt{3}i+4}{3-1}=\dfrac{-8+8\sqrt{3}i}{2}=-4+4\sqrt{3}i\end{array}}$