Verified answer

[tex]\displaystyle \sf (1+2i)(1+3i) \to -5+5i \\\\\ \text{Seja o complexo} : \\\\ z = a+b\cdot i \\\\ \theta =arg(z) \to tg(\theta) = \frac{b}{a} \to arg(z) = Arc\ tg\left(\frac{b}{a}\right) \\\\\ \text{Sejam os complexos } : \\\\ z_1 = cis(\theta) \to arg(z_1) = \theta \\\\ z_2 =cis(\alpha) \to arg(z_2)=\alpha \\\\ z_1\cdot z_2 = cis(\theta+\alpha) \to arg(z_1+z_2) = \theta+\beta \to \boxed{\sf arg(z_1\cdot z_2) = arg(z_1)+arg(z_2) }[/tex]

Façamos :

[tex]\displaystyle \sf z_ 1 = 1+2i \to 0 < arg(z_1) < \frac{\pi}{2} \to \text{Parte real positiva e parte imagin\'aria positiva } \\\\ arg(z_1) = Arc \ tg\left(\frac{2}{1}\right) \to arg(z_1) = Arc\ tg(2) \\\\\\ z_ 2 = 1+3i \to 0 < arg(z_2) < \frac{\pi }{2} \to \text{Parte real positiva e parte imagin\'aria positiva } \\\\\ arg(z_2) = Arc \ tg\left(\frac{3}{1}\right) \to arg(z_2) = Arc\ tg(3)[/tex]

[tex]\displaystyle \sf \boxed{\sf 0 < arg(z_1)+arg(z_2) < \pi } \\\\\\ arg(z_1)+arg(z_2) = Arc\ tg(2)+Arc\ tg(3) \\\\ arg(z_1\cdot z_2) = Arc \ tg(2)+Arc\ tg(3) \\\\\\ \boxed{\sf arg(z_1\cdot z_2) = arg(-5+5i) = Arc\ tg\left(\frac{5}{-5}\right) = Arc\ tg(-1) } \\\\\ Da{\'i}}: \\\\ Arc tg(-1) = Arc \ tg(2)+Arc\ tg(3)[/tex]

[tex]\displaystyle \sf \Large\boxed{\sf \ \frac{3\pi}{4} = Arc \ tg(2)+Arc\ tg(3)\ }\checkmark[/tex]