Найти f(z0) если f(z)=z+2/z(z-1) и z0=1+2i

[latex]f( z_{0})=(1+2i)+ \frac{2}{(1+2i)(1+2i-1)}=1+2i+ \frac{2}{(1+2i)2i}=1+2i+ \frac{1}{i+2 i^{2} }=1+2i+ \frac{1}{i-2}=1+2i- \frac{1}{2-i}= \frac{(1+2i)(2-i)-1}{2-i}= \frac{2+4i-i- 2i^{2}-1}{2-i}= \frac{3+3i}{2-i}= \frac{3(1+i)(2+i)}{(2-i)(2+i)}= \frac{3(2+2i+i+ i^{2})}{4- i^{2} } = \frac{3(1+3i)}{4+1}= \frac{3+9i}{5} [/latex] по определению мнимой единицы: [latex] i^{2}=-1 [/latex]