a) Como
\n$$\\begin{array}{*{20}{l}}
\n{f(z)}& = &{\\frac{4}{{x + yi}} + 1 + i} \\\\
\n{}& = &{\\frac{4}{{x + yi}} \\times \\frac{{x – yi}}{{x – yi}} + 1 + i} \\\\
\n{}& = &{\\frac{{4x – 4yi}}{{{x^2} + {y^2}}} + 1 + i} \\\\
\n{}& = &{\\frac{{\\left( {{x^2} + {y^2} + 4x} \\right) + \\left( {{x^2} + {y^2} – 4y} \\right)i}}{{{x^2} + {y^2}}}} \\\\
\n{}& = &{\\frac{{{x^2} + {y^2} + 4x}}{{{x^2} + {y^2}}} + \\frac{{{x^2} + {y^2} – 4y}}{{{x^2} + {y^2}}}i}
\n\\end{array}$$
\nent\u00e3o $$\\begin{array}{*{20}{l}}
\nX& = &{\\operatorname{Re} \\left( {f(z)} \\right)} \\\\
\n{}& = &{\\frac{{{x^2} + {y^2} + 4x}}{{{x^2} + {y^2}}}}
\n\\end{array}$$ e $$\\begin{array}{*{20}{l}}
\nY& = &{\\operatorname{Im} \\left( {f(z)} \\right)} \\\\
\n{}& = &{\\frac{{{x^2} + {y^2} – 4y}}{{{x^2} + {y^2}}}}
\n\\end{array}$$<\/p>\nb) Para que ${f(z)}$ seja um n\u00famero real, ter\u00e1 de ser $Y = \\operatorname{Im} \\left( {f(z)} \\right) = 0$: $$\\begin{array}{*{20}{l}}
\n{Y = 0}& \\Leftrightarrow &{\\frac{{{x^2} + {y^2} – 4y}}{{{x^2} + {y^2}}} = 0} \\\\
\n{}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{{x^2} + {y^2} – 4y = 0}& \\wedge &{{x^2} + {y^2} \\ne 0}
\n\\end{array}} \\\\
\n{}& \\Leftrightarrow &{\\begin{array}{*{20}{l}}
\n{{x^2} + {{\\left( {y – 2} \\right)}^2} = 4}& \\wedge &{{x^2} + {y^2} \\ne 0}
\n\\end{array}}
\n\\end{array}$$
\nNota: Como $z \\ne 0 + 0i$, ent\u00e3o ${x^2} + {y^2} \\ne 0$.<\/p>\n
O conjunto F dos pontos M afixos de $z$ tais que $f(z)$ seja um n\u00famero real \u00e9 a circunfer\u00eancia de centro $\\left( {0,2} \\right)$ e raio 2 unidades, com exce\u00e7\u00e3o do ponto de coordenadas $\\left( {0,0} \\right)$.<\/p>\n
c) O afixo da solu\u00e7\u00e3o da equa\u00e7\u00e3o $f(z) = 4$ pertence a F, pois as suas coordenadas verificam a equa\u00e7\u00e3o que define F: $$\\begin{array}{*{20}{l}}
\n{{{\\left( {\\frac{6}{5}} \\right)}^2} + {{\\left( {\\frac{2}{5} – 2} \\right)}^2} = 4}& \\Leftrightarrow &{\\frac{{36}}{{25}} + \\frac{{64}}{{25}} = 4} \\\\
\n{}& \\Leftrightarrow &{4 = 4}
\n\\end{array}$$<\/p>\n<\/li>\n<\/ol>\n