(i) z+zˉ2=Re(z)\frac{z+\bar{z}}{2}=\operatorname{Re}(z)2z+zˉ=Re(z) Solution Let z=a+ib, a,b∈R.Then zˉ=a−ib.z+zˉ=(a+ib)+(a−ib)=2az+zˉ2=2a2=a=Re(z)\begin{aligned} & \boxed{\text{Let } z=a+ib,\ a,b\in\mathbb{R}.} \\ \\ & \boxed{\text{Then } \bar{z}=a-ib.} \\ \\ & z+\bar{z}=(a+ib)+(a-ib)=2a \\ \\ & \frac{z+\bar{z}}{2}=\frac{2a}{2}=a=\operatorname{Re}(z) \end{aligned}Let z=a+ib, a,b∈R.Then zˉ=a−ib.z+zˉ=(a+ib)+(a−ib)=2a2z+zˉ=22a=a=Re(z)