Question

2) calcule o módulo dos seguintes números complexos :
3)ache o módulo dos números complexos :​

Answer 1

$\large\boxed{\begin{array}{l}\rm 2)~\sf z=4-i\\\sf \rho=\sqrt{4^2+(-1)^2}=\\\sf \rho=\sqrt{16+1}\\\sf\rho=\sqrt{17}\\\rm b)~\sf z=-5i\\\sf \rho=\sqrt{(-5)^2}=|-5|=5\\\rm c)~\sf z=\sqrt{2}+i\sqrt{3}\\\sf \rho=\sqrt{(\sqrt{2})^2+(\sqrt{3})^2}\\\sf \rho=\sqrt{2+3}\\\sf\rho=\sqrt{5}\\\rm d)~\sf z=\dfrac{1}{2}+\dfrac{1}{3}i\\\sf \rho=\sqrt{\bigg(\dfrac{1}{2}\bigg)^2+\bigg(\dfrac{1}{3}\bigg)^2}\\\\\sf\rho=\sqrt{\dfrac{1}{4}+\dfrac{1}{9}}\\\\\sf\rho=\sqrt{\dfrac{9+4}{36}}=\dfrac{\sqrt{13}}{6}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm e)~\sf z=8\\\sf \rho=\sqrt{8^2}=|8|=8\\\rm f)~\sf z=0\\\sf \rho=\sqrt{0^2}=|0|=0\end{array}}$

$\Large\boxed{\begin{array}{l}\rm 3)\\\rm a)~\sf z=(3-i)(2+i)\\\sf z=6+3i-2i-i^2\\\sf z=6+3i-2i+1\\\sf z=7+i\\\sf \rho=\sqrt{7^2+1^2}\\\sf \rho=\sqrt{49+1}\\\sf \rho=\sqrt{50}=\sqrt{25\cdot2}\\\sf \rho=5\sqrt{2}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm b)~\sf z=\dfrac{(1+4i)}{i}\cdot\dfrac{i}{i}\\\\\sf z=\dfrac{i+4i^2}{i^2}\\\\\sf z=\dfrac{i+4\cdot(-1)}{-1}\\\sf z=4-i\\\sf \rho=\sqrt{4^2+(-1)^2}\\\sf\rho=\sqrt{16+1}\\\sf \rho=\sqrt{17}\end{array}}$

$\Large\boxed{\begin{array}{l}\rm c)~\sf z=\dfrac{(4-3i)(12-5i)}{\sqrt{2}i}\cdot\dfrac{\sqrt{2}i}{\sqrt{2}i}\\\\\sf z=\dfrac{[48-20i-36i+15i^2]\cdot\sqrt{2}i}{2i^2}\\\\\sf z=\dfrac{[48-20i-36i-15]\cdot\sqrt{2}i}{-2}\\\\\sf z=\dfrac{[33-56i]\cdot\sqrt{2}i}{-2}\\\\\sf z=\dfrac{33\sqrt{2}i-56\sqrt{2}i^2}{-2}\\\\\sf z=\dfrac{33\sqrt{2}i+56\sqrt{2}}{-2}\\\\\sf z=-28\sqrt{2}-\dfrac{33\sqrt{2}}{2}i\end{array}}$

$\Large\boxed{\begin{array}{l}\sf \rho=\sqrt{(-28\sqrt{2})^2+\bigg(-\dfrac{33\sqrt{2}}{2}\bigg)^2}\\\\\sf \rho=\sqrt{784\cdot2+\dfrac{1089\cdot\backslash\!\!\!2}{\backslash\!\!\!\!4}}\\\\\sf \rho=\sqrt{1568+\dfrac{1089}{2}}\\\\\sf \rho=\sqrt{\dfrac{3136+1089}{2}}\\\\\sf \rho=\sqrt{\dfrac{4225}{2}}\\\\\sf \rho=\dfrac{\sqrt{4225}}{\sqrt{2}}\cdot\dfrac{\sqrt{2}}{\sqrt{2}}\\\\\sf \rho=\dfrac{65\sqrt{2}}{2}\end{array}}$