Question

(Mackenzie 1996) Se k é um número real e o argumento de z = (k + 2i)/(3 - 2i) é π/4, então pertence ao intervalo:? heeelllpppp :)

Answer 1

$\mathrm{Z=\dfrac{k+2i}{3-2i}\ \to\ Z=\dfrac{k+2i}{3-2i}.\dfrac{3+2i}{3+2i}}\\\\\\ \mathrm{Z=\dfrac{3k+2ki+6i+4i^2}{9+6i-6i-4i^2}\ \to\ Z=\dfrac{3k+2ki+6i+4(-1)}{9-4(-1)}}\\\\\\ \mathrm{Z=\dfrac{(3k-4)+(2k+6)i}{13}\ \to\ \boxed{\mathrm{Z=\bigg(\dfrac{3k-4}{13}\bigg)+\bigg(\dfrac{2k+6}{13}\bigg)i}}}$

$\mathrm{\Rightarrow Se\ \arg{(Z)}=\theta=\dfrac{\pi}{4},\ \sin{\theta}=\cos{\theta}.\ Logo:}\\\\\\ \mathrm{\dfrac{b}{|Z|}=\dfrac{a}{|Z|}\ \to\ a=b\ \to\ \dfrac{3k-4}{13}=\dfrac{2k+6}{13}}\\\\\\ \mathrm{3k-4=2k+6\ \to\ 3k-2k=6+4\ \to\ \boxed{\boxed{\mathbf{k=10}}}}$