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# determine o argumento Θ do numero complexo z nos seguintes casos a)z=(1,1) b)z=-√3+i c)z=√2-i d)z=4+4i e)z=-6i f)z=-12

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$\mathrm{\mathbf{a)}\ Z=1+i\ \to\ Z=(1,1)\ \to\ Z\in1\ºQ.}\\\\ \mathrm{\tan{\theta}=\dfrac{b}{a}\ \to\ \tan{\theta}=\dfrac{1}{1}\ \to\ \tan{\theta=1}}\\\\ \mathrm{\theta=\arctan{(1)}\ \to\ \boxed{\mathrm{\theta=45\º=\dfrac{\pi}{4}}}}$ $\mathrm{\mathbf{b)}\ Z=-\sqrt{3}+i\ \to\ Z=(-\sqrt{3},1)\ \to\ Z\in2\ºQ.}\\\\ \mathrm{\tan{\theta}=\dfrac{b}{a}\ \to\ \tan{\theta}=\dfrac{1}{-\sqrt{3}}\ \to\ \tan{\theta}=\dfrac{-\sqrt{3}}{3}}\\\\ \mathrm{\theta=\arctan{\bigg(\dfrac{-\sqrt{3}}{3}\bigg)}\ \to\ \boxed{\mathrm{\theta=150\º=\dfrac{5\pi}{6}}}}$ $\mathrm{\mathbf{c)}\ Z=\sqrt{2}-i\ \to\ Z=(\sqrt{2},-1)\ \to\ Z\in4\ºQ.}\\\\ \mathrm{\tan{\theta}=\dfrac{b}{a}\ \to\ \tan{\theta}=\dfrac{-1}{\sqrt{2}}\ \to\ \tan{\theta}=\dfrac{-\sqrt{2}}{2}}\\\\ \boxed{\mathrm{\theta=\arctan{\bigg(\dfrac{-\sqrt{2}}{2}\bigg)}}}$ $\mathrm{\mathbf{d)}\ Z=4+4i\ \to\ Z=(4,4)\ \to\ Z\in1\ºQ.}\\\\ \mathrm{\tan{\theta}=\dfrac{b}{a}\ \to\ \tan{\theta}=\dfrac{4}{4}\ \to\ \tan{\theta}=1}\\\\ \mathrm{\theta}=\arctan{(1)}\ \to\ \boxed{\mathrm{\theta=45\º=\dfrac{\pi}{4}}}}$ $\mathrm{\mathbf{e)}\ Z=-6i\ \to\ Z=(0,-6)}\\\\ \mathrm{|Z|=\sqrt{a^2+b^2}=\sqrt{0^2+(-6)^2}\ \to\ |Z|=6}\\\\ \mathrm{\sin{\theta}=\dfrac{b}{|Z|}\ \to\ \sin{\theta}\ \to\ \sin{\theta}=\dfrac{-6}{6}\ \to\ \sin{\theta}=-1}\\\\ \mathrm{\theta=\arcsin{(-1)}\ \to\ \boxed{\mathrm{\theta=270\º=\dfrac{3\pi}{2}}}}$ $\mathrm{\mathbf{f)}\ Z=-12\ \to\ Z=(-12,0)}\\\\ \mathrm{|Z|=\sqrt{a^2+b^2}=\sqrt{(-12)^2+0^2}\ \to\ |Z|=12}\\\\ \mathrm{\cos{\theta}=\dfrac{a}{|Z|}\ \to\ \cos{\theta}=\dfrac{-12}{12}\ \to\ \cos{\theta}=-1}\\\\ \mathrm{\theta=\arccos{(-1)}\ \to\ \boxed{\mathrm{\theta=180\º=\pi}}}$