[tex]\displaystyle \sf z=a+b\ i\ ;\\\\ |z|=\sqrt{a^2+b^2} \\\\ temos : \\\\ |z| = |1-z|=\left|\frac{1}{z}\right| \\\\\ |a+b\ i|=|1-(a+b\ i)| = \left|\frac{1}{a+b\ i}\right| \\\\\\ \sqrt{a^2+b^2} = |1-a-b\ i|=\frac{1}{\sqrt{a^2+b^2}} \\\\\\\ \sqrt{a^2+b^2}=\sqrt{(1-a)^2+(-b)^2} = \left(\sqrt{a^2+b^2}\right)^{-1}\\\\\\ a^2+b^2=(1-a)^2+b^2=\left(a^2+b^2\right)^{-1} \\\\\ a^2+b^2=1-2a+a^2+b^2 \\\\ a^2+b^2-a^2-b^2=1-2a \\\\ 1-2a = 0 \\\\ 2a = 1\\\\ \boxed{\sf a = \frac{1}{2} }[/tex]
[tex]\displaystyle \sf \text{Fa\c ca} : \\\\ a^2+b^2= \left(a^2+b^2\right)^{-1} \\\\\ a^2+b^2=\frac{1}{a^2+b^2} \\\\\\ \left(a^2+b^2\right)\cdot \left(a^2+b^2\right)=1\\\\ \left(a^2+b^2\right)^2= 1 \\\\ \text{m\'odulo \'e positivo, ent\~ao n precisa se precupar com} \pm aqui.\\\\ a^2+b^2 = 1 \\\\ \left(\frac{1}{2}\right)^2+b^2=1 \\\\ \frac{1}{4}+b^2=1\\\\\ b^2=1-\frac{1}{4}\\\\\ b^2=\frac{4-1}{4}=\frac{3}{4}\\\\\ b=\pm \sqrt{\frac{3}{4}} \\\\\\ b = \pm \frac{\sqrt{3}}{2}[/tex]
[tex]\displaystyle \sf \text{a quest\~ao diz que }: \\\\ z = a+b\ i\\\\ ent\~ao : \\\\ \large\boxed{\sf z =\frac{1}{2}+\frac{\sqrt{3}}{2}\ i }\checkmark[/tex]
por curiosidade, testando o z :
[tex]\displaystyle \sf z=\frac{1}{2}+\frac{\sqrt{3}}{2}\ i\\\\ |z|=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} \\\\\\ |z| = \sqrt{\frac{1}{4}+\frac{3}{4}} =\sqrt{\frac{4}{4}}=\sqrt{1}=1\\\\|z|= 1\\\\ |z| = |1-z|=\left|\frac{1}{z}\right| \\\\\\\ 1 = \left|1-\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\ i\right)\right|=\frac{1}{1} \\\\\\\ 1 = \left|\left(1-\frac{1}{2}\right)-\frac{\sqrt{3}}{2}\ i\right| = 1[/tex]
[tex]\displaystyle \sf 1=\left|\frac{1}{2}-\frac{\sqrt{3}}{2}\ i\right|=1\\\\\\\ 1=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{-\sqrt{3}}{2}\right)^2} = 1 \\\\\\\ 1 = 1 = 1 \checkmark[/tex]
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[tex]\displaystyle \sf z=a+b\ i\ ;\\\\ |z|=\sqrt{a^2+b^2} \\\\ temos : \\\\ |z| = |1-z|=\left|\frac{1}{z}\right| \\\\\ |a+b\ i|=|1-(a+b\ i)| = \left|\frac{1}{a+b\ i}\right| \\\\\\ \sqrt{a^2+b^2} = |1-a-b\ i|=\frac{1}{\sqrt{a^2+b^2}} \\\\\\\ \sqrt{a^2+b^2}=\sqrt{(1-a)^2+(-b)^2} = \left(\sqrt{a^2+b^2}\right)^{-1}\\\\\\ a^2+b^2=(1-a)^2+b^2=\left(a^2+b^2\right)^{-1} \\\\\ a^2+b^2=1-2a+a^2+b^2 \\\\ a^2+b^2-a^2-b^2=1-2a \\\\ 1-2a = 0 \\\\ 2a = 1\\\\ \boxed{\sf a = \frac{1}{2} }[/tex]
[tex]\displaystyle \sf \text{Fa\c ca} : \\\\ a^2+b^2= \left(a^2+b^2\right)^{-1} \\\\\ a^2+b^2=\frac{1}{a^2+b^2} \\\\\\ \left(a^2+b^2\right)\cdot \left(a^2+b^2\right)=1\\\\ \left(a^2+b^2\right)^2= 1 \\\\ \text{m\'odulo \'e positivo, ent\~ao n precisa se precupar com} \pm aqui.\\\\ a^2+b^2 = 1 \\\\ \left(\frac{1}{2}\right)^2+b^2=1 \\\\ \frac{1}{4}+b^2=1\\\\\ b^2=1-\frac{1}{4}\\\\\ b^2=\frac{4-1}{4}=\frac{3}{4}\\\\\ b=\pm \sqrt{\frac{3}{4}} \\\\\\ b = \pm \frac{\sqrt{3}}{2}[/tex]
[tex]\displaystyle \sf \text{a quest\~ao diz que }: \\\\ z = a+b\ i\\\\ ent\~ao : \\\\ \large\boxed{\sf z =\frac{1}{2}+\frac{\sqrt{3}}{2}\ i }\checkmark[/tex]
por curiosidade, testando o z :
[tex]\displaystyle \sf z=\frac{1}{2}+\frac{\sqrt{3}}{2}\ i\\\\ |z|=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} \\\\\\ |z| = \sqrt{\frac{1}{4}+\frac{3}{4}} =\sqrt{\frac{4}{4}}=\sqrt{1}=1\\\\|z|= 1\\\\ |z| = |1-z|=\left|\frac{1}{z}\right| \\\\\\\ 1 = \left|1-\left(\frac{1}{2}+\frac{\sqrt{3}}{2}\ i\right)\right|=\frac{1}{1} \\\\\\\ 1 = \left|\left(1-\frac{1}{2}\right)-\frac{\sqrt{3}}{2}\ i\right| = 1[/tex]
[tex]\displaystyle \sf 1=\left|\frac{1}{2}-\frac{\sqrt{3}}{2}\ i\right|=1\\\\\\\ 1=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{-\sqrt{3}}{2}\right)^2} = 1 \\\\\\\ 1 = 1 = 1 \checkmark[/tex]