Question

4)efetue :
5) efetue :
6)efetue as divisões indicadas :
7)no anexo ​

Answer 1

$\Large\boxed{\begin{array}{l}\boldsymbol{4}\\\rm a)~\sf (5+2i)\cdot(2-i)=10-5i+4i-2i^2\\\sf(5+2i)\cdot(2-i)=10-i+2=12-i\\\rm b)~\sf(-1+2i)(3+i)=-3-i+6i+3i^2\\\sf(-1+2i)\cdot(3+i)=-3+5i-3=5i-6\\\rm c)~\sf\bigg(\dfrac{1}{2}+i\bigg)\cdot\bigg(\dfrac{1}{2}-i\bigg)=\bigg(\dfrac{1}{2}\bigg)^2-i^2\\\\\sf\bigg(\dfrac{1}{2}-i\bigg)\cdot\bigg(\dfrac{1}{2}+i\bigg)=\dfrac{1}{4}+1=\dfrac{5}{4}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm d)~\sf(-3i+4)\cdot(-2+5)=3\cdot(-3i+4)=-9i+12\end{array}}$

$\Large\boxed{\begin{array}{l}\boldsymbol{5}\\\rm a)~\sf(3+4i)^2=9+24i-16=-7+24i\\\rm b)~\sf(2+i)^3=2+11i\end{array}}$

$\Large\boxed{\begin{array}{l}\boldsymbol{6}\\\rm a)~\sf\dfrac{(2+3i)}{(1+2i)}\cdot\dfrac{(1-2i)}{(1-2i)},\\\\\sf\dfrac{8-i}{2^2-(3i)^2}\\\\\sf\dfrac{8-i}{4+9}=\dfrac{8}{13}-\dfrac{1}{13}i\end{array}}$

$\Large\boxed{\begin{array}{l}\rm b)~\sf\dfrac{1}{(3+2i)}\cdot\dfrac{(3-2i)}{(3-2i)}\\\\\sf\dfrac{3-2i}{3^2-(-2i)^2}\\\\\sf\dfrac{3-2i}{9+4}=\dfrac{3}{13}-\dfrac{2}{13}i\end{array}}$

$\Large\boxed{\begin{array}{l}\boldsymbol{7}\\\sf Z=i^{101}+i^{102}+i^{103}+i^{104}+i^{105}+i^{106}\\\sf i^{101}=i^{1}=i\\\sf i^{102}=i^2=-1\\\sf i^{103}=i^{3}=-i\\\sf i^{104}=i^4=1\\\sf i^{105}=i^1=i\\\sf i^{106}=i^2=-1\\\underline{\rm substituindo~temos:}\\\sf Z=i+(-1)+(-i)+1+i+(-1)\\\sf Z=\backslash\!\!\!i-1-\backslash\!\!\!i+\backslash\!\!\!1+i-\backslash\!\!\!1\\\sf Z=i-1\end{array}}$