根据材料回答：
(1) 填空：$i^{4}=$
(2)$(a + bi)^{2}$的运算符合完全平方公式，求$(2 + 3i)^{2}$的值；
(3)已知$(a + i)(b + i)=2 - 5i$，求$(a^{2}+b^{2})(i^{2}+i^{3}+i^{4}+··· +i^{2026})$的值。
(1) 填空：$i^{4}=$
1
，$i^{2}+i^{3}+i^{4}+i^{5}=$0
；(2)$(a + bi)^{2}$的运算符合完全平方公式，求$(2 + 3i)^{2}$的值；
(3)已知$(a + i)(b + i)=2 - 5i$，求$(a^{2}+b^{2})(i^{2}+i^{3}+i^{4}+··· +i^{2026})$的值。
答案：(1) 1 0 解析:$\mathrm{i}^{4}=(-1)×(-1)=1$,$\mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}+\mathrm{i}^{5}=-1+(-\mathrm{i})+1+\mathrm{i}=0$.
(2) $(2 + 3\mathrm{i})^{2}=2^{2}+12\mathrm{i}+9\mathrm{i}^{2}=4 + 12\mathrm{i}-9=-5 + 12\mathrm{i}$.
(3) $\because (a + \mathrm{i})(b + \mathrm{i})=ab - 1+(a + b)\mathrm{i}=2 - 5\mathrm{i}$,$\therefore ab - 1 = 2$,$a + b=-5$,即$ab = 3$,$a + b=-5$.又$\because \mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}+\mathrm{i}^{5}=-1-\mathrm{i}+1+\mathrm{i}=0$,$\mathrm{i}^{6}+\mathrm{i}^{7}+\mathrm{i}^{8}+\mathrm{i}^{9}=-1-\mathrm{i}+1+\mathrm{i}=0$,故4个一组为一个循环.$\because (2026 - 1)÷4 = 506······1$,$\therefore (a^{2}+b^{2})(\mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}+···+\mathrm{i}^{2026})=[(a + b)^{2}-2ab](506×0 - 1)=[(-5)^{2}-2×3]×(-1)=-19$.
(2) $(2 + 3\mathrm{i})^{2}=2^{2}+12\mathrm{i}+9\mathrm{i}^{2}=4 + 12\mathrm{i}-9=-5 + 12\mathrm{i}$.
(3) $\because (a + \mathrm{i})(b + \mathrm{i})=ab - 1+(a + b)\mathrm{i}=2 - 5\mathrm{i}$,$\therefore ab - 1 = 2$,$a + b=-5$,即$ab = 3$,$a + b=-5$.又$\because \mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}+\mathrm{i}^{5}=-1-\mathrm{i}+1+\mathrm{i}=0$,$\mathrm{i}^{6}+\mathrm{i}^{7}+\mathrm{i}^{8}+\mathrm{i}^{9}=-1-\mathrm{i}+1+\mathrm{i}=0$,故4个一组为一个循环.$\because (2026 - 1)÷4 = 506······1$,$\therefore (a^{2}+b^{2})(\mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}+···+\mathrm{i}^{2026})=[(a + b)^{2}-2ab](506×0 - 1)=[(-5)^{2}-2×3]×(-1)=-19$.