For 𝑓(𝑧)=1(𝑧−𝑖)(𝑧+𝑖) consider 𝑓(𝑧)=𝐴𝑧−𝑖+𝐵𝑧+𝑖 and solve 𝐴,𝐵 And it’s 𝑓(𝑧)=−𝑖/2𝑧−𝑖+𝑖/2𝑧+𝑖. Then do Power theory: 𝑓(𝑧)=12(11−(−𝑧𝑖)+11−𝑧𝑖)=12[∑∞𝑘=0(−𝑧𝑖)𝑘+∑∞𝑘=0(𝑧𝑖)𝑘]=∑∞𝑘=0𝑧2𝑘𝑖2𝑘 □