Question

Calcule o módulo do número complexo.

Answer 1

Módulo de um número complexo :

$\displaystyle \sf z = a+i\cdot b \\\\ m{\'odulo}: \\\\ |z|=\sqrt{a^2+b^2 }$

temos :

$\displaystyle \sf z= \frac{1}{1+i\cdot tg(x) } \\\\\\ racionalizando : \\\\z= \frac{1}{[1+i\cdot tg(x)] }\cdot \frac{\left[1-i\cdot tg(x) \right] }{\left[1-i\cdot tg(x) \right] } \\\\\\ z= \frac{1-i\cdot tg(x)}{1-i^2\cdot tg^2(x) } \to \frac{1-i\cdot tg(x)}{1-(-1)\cdot tg^2(x)} \\\\\\ z=\frac{1-i\cdot tg(x) }{1+tg^2(x)} \\\\\\ z= \frac{1}{\underbrace{\sf1+tg^2(x)}_{sec^2(x)}}-\frac{i\cdot tg(x)}{\underbrace{\sf1+tg^2(x)}_{sec^2(x)}} \\\\\\ z= \frac{1}{sec^2(x)}-\frac{i\cdot tg(x)}{sec^2(x)}$

Daí :

$\displaystyle \sf |z| = \sqrt{\left(\frac{1}{sec^2(x)}\right)^2+\left(\frac{tg(x)}{sec^2(x)}\right)^2} \\\\\\ |z|= \sqrt{\frac{1}{sec^4(x)}+\frac{tg^2(x)}{sec^4(x)}} \\\\\\ |z|=\sqrt{\frac{1+tg^2(x)}{sec^4(x)}} = \sqrt{\frac{sec^2(x)}{sec^4(x)} } \\\\\\ |z|= \sqrt{\frac{1}{sec^2(x)}} \\\\\\\ |z|=\frac{1}{sec(x)} \\\\\\\ obs : \ \frac{1}{cos(x)}=sec(x) \ \to \ \frac{1}{sec(x)} = cos(x) \\\\ Portanto : \\\\ \huge\boxed{\sf\ |z| = cos(x) \ }\checkmark$