Originally Posted by

**bkbowser** Hello,

The problem is: Define $\text{Sqrt}(z)=|z|^{1/2}e^{i \theta /2}$ where $z \in \mathbb{C} \backslash (-\infty,0]$ and $-\pi < \theta <\pi$.

a) Find two numbers $z_1,z_2 \in \mathbb{C} \backslash (-\infty,0]$ such that $w=z_1z_2 \in \mathbb{C} \backslash (-\infty,0]$ but $\text{Sqrt}{(w)}\neq \text{Sqrt}{(z_1)}\text{Sqrt}{(z_2)}$.

b) Find the subset $S \subseteq \mathbb{C} \backslash (-\infty,0]$ such that if $z \in S$ then $\text{Sqrt}{(iz)}=\text{Sqrt}{(i)}\text{Sqrt}{(z) }$.

For part a) I think $2i$ and $-2i$ work (but I've screwed this up before). We'd have $\text{Sqrt}{(2i)}=\sqrt{2}e^{i\pi/4}$, $\text{Sqrt}{(-2i)}=\sqrt{2}e^{i \pi 3/4}$, and $\text{Sqrt}{(2i(-2i))}=\text{Sqrt}{(4)}=\sqrt{4}e^{i 0}$. Then $\sqrt{2}e^{i\pi/4}\sqrt{2}e^{i \pi 3/4} \neq \sqrt{4}e^{i 0}$.