The argument of $$\frac{1+i \sqrt{3}}{\sqrt{3}+i}, i=\sqrt{-1}$$ is

If $$w=\frac{z}{z-\frac{1}{3} i}$$ and $$|w|=1, i=\sqrt{-1}$$, then $$z$$ lies on

If $$Z_1=2+i$$ and $$Z_2=3-4 i$$ and $$\frac{\overline{Z_1}}{Z_2}=a+b i$$, then the value of $$-7 a+b$$ is (where $$i=\sqrt{-1}$$ and $$a, b \in R)$$

If $$Z_1=4 i^{40}-5 i^{35}+6 i^{17}+2, Z_2=-1+i$$, where $$i=\sqrt{-1}$$, then $$\left|Z_1+Z_2\right|=$$