Fie z=a+bi, I z I=[tex] \sqrt{ a^{2}+b^2 } [/tex], si I z-1 I=I (a-1)+bi I=[tex] \sqrt{(a-1)^2+b^2} [/tex], deci avem: [tex] \sqrt{(a-1)^2+b^2}+i \sqrt{a^2+b^2}=1+i [/tex]. Egaland partile reale intre ele si coeficentii partilor imaginare intre ei, avem ambi radicali egali cu 1 ii egalam si ridicand egalitatea la patrat obtinem: [tex] a^{2} + b^{2}= a^{2}-2a+1+b^2, [/tex], rezulta a=[tex] \frac{1}{2},inlocuind.in. a^{2} + b^{2}=1.rezulta.y^2= \frac{3}{4},deci. y_{1}=- \frac{ \sqrt{3} }{2}.si. y_{2}= \frac{ \sqrt{3} }{2} [/tex], deci avem doua solutii:
[tex] z_{1}= \frac{1}{2}-i \frac{ \sqrt{3} }{2},si. z_{2}= \frac{1}{2}+ i\frac{ \sqrt{3} }{2}. [/tex]
