Given some $f(z)$ which is bounded by the open ball $B(1,1)$ in the complex plane, is $\arg(f(z))$ continuous? It says it is, but I do not see why. For instance inside the open ball $B(1,1)$ we can have some closed ball $\overline {B(1,0.5)}$ which is defined by the function $0.5e^{iz}$, now if we have $\arg(0.5e^{i0})=\arg(0.5)=2\pi$ and $\arg(0.5e^{i2\pi})=\arg(0.5)=2\pi+2k\pi$, but clearly they are not equal so they are not continuous, and hence $\arg (f(z)) \in B(1,1)$ cannot be continuous?
Am I correct, or am I failing to see something here?
For $z\in B(1,1)$, we have $\text{Re}(z)> 0$, hence, the principal value of $\text{arg}(z)$ is strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, and there are no jumps.
Explicitly, for $z\in B(1,1)$, if we write $z=x+iy$, we always have $x>0$, hence the principal value of $\text{arg}(z)$ is given by $$\text{arg}(z)=\tan^{-1}\left(\frac{y}{x}\right)$$ Then since $x,y$ are continuous functions of $z$ (and $x\ne 0)$, it follows (assuming principal values) that $\text{arg}(z)$ is continuous on $B(1,1)$.
Note that for an arbitrary continuous function $f:\mathbb{C}\to \mathbb{C}$, the function $\text{arg}(f(z))$ need not be continuous.
For example, if the image of $f$ contains an open ball centered at $0$, then for that function $f$, no choice of interval for $\text{arg}$ can make $\arg(f(z))$ continuous.
But for this problem, it's given that $f:\mathbb{C}\to C$ is such that the image of $f$ is contained in $B(1,1)$. Also, I'll assume the actual problem required $f$ to be continuous.
Thus, since $f$ is continuous on $\mathbb{C}$, and $\text{arg}(z)$ is continuous on the image of $f$, it follows that $\text{arg}(f(z))$ is continuous.