,

$$\begin{eqnarray*} \frac{1}{1+\frac{1}{z}} & = & \frac{1}{\frac{z}{z}+\frac{1}{z}}\\ & = & \frac{z}{z+1} \end{eqnarray*}$$

,

จะได้ผลรวมของทั้งสี่เทอมเป็น

$$\begin{array}{l}  \frac{1+i}{(1+i)+1}+\frac{1-i}{(1-i)+1}+\frac{-1+i}{(-1+i)+1}+\frac{1+i}{(1+i)+1}\\ \qquad = \frac{1+i}{2+i}+\frac{1-i}{2-i}+\frac{-1+i}{i}+\frac{1+i}{2+i}\\ \qquad = 2\frac{(1+i)(2-i)}{5}+\frac{(1-i)(2+i)}{5}+\frac{(-1+i)(-i)}{1}\\ \qquad = \frac{2}{5}(3+i)+\frac{1}{5}(3-i)+1+i\\ \qquad = \frac{6+3+5}{5}+\frac{2+1+5}{5}i\\ \qquad = \frac{14}{5}+\frac{6}{5}i \end{array}$$

ดังนั้น $a=\frac{14}{5}$ และ $b=\frac{6}{5}$