How do I graph 0 < Arg(z - i) < pi / 4? How do I go about graphing the region $0 < Arg(z- i) <$ $\frac{\pi}{4}$.
 A: Excuse the terrible images. They're supposed to be Argand diagrams (in the complex plane). 
Construct a preliminary diagram. Start by constructing rays representing arguments of $0$ (horizontal ray stretching to the right to infinity) and $\frac{\pi}{4}$ (a ray bisecting the first quadrant stretching upward and rightward to infinity). The blue area in between (strictly excluding the two rays themselves) represents the locus of points satisfying $0 < \arg(w) < \frac{\pi}{4}$.
This gives this image for the locus of $w$:

To get the locus of $z$, add $i$ to $w$, and this involves a vertical upward translation of $1$ unit. That gives you the following image, that represents the locus of $z$, where $0 < \arg(z-i) < \frac{\pi}{4}$:

A: Lets divide $0<\arg(z-i)<\pi/4$ into:
$$\begin{cases}0\lt\arg(a+ib-i)\\\arg(a+bi-i)=\arg(a+i(b-1))\lt\pi/4\end{cases}$$
Now we need to look of what value of $a$ and $b$ both of them are true.
It is easy to see that $a$ is positive and $b$ is greater than $1$. 
For what value of $b$ we get $\arg(c+id)=\pi/4$? When $c=d$, so when $a=b-1\implies a+1=b$ now for fixed $a$ we have $\arg(a+ib-i)$ increasing when $b$ increasing so the answer for the second case is $b<a+1$
Draw both cases on the same graph and where they both exists there is the graph you are looking for
A: $$0 \lt Arg(z-i) \lt \frac{\pi}{4}$$
replace $z=x+(yi)$:
$$0 \lt Arg(x+(yi)-i) \lt \frac{\pi}{4}$$
$$0 \lt Arg(x+(y-1)i) \lt \frac{\pi}{4}$$
Definition of $Arg(z=x+(y-1)i)=atan2(x,y-1)$ (see definition of atan2 here):
$$0 \lt atan2(x,y-1) \lt \frac{\pi}{4}$$
We know by the initial expression that the argument is inside the interval $[0, \frac{\pi}{4}]$. It means that both $x$ and $y-1$ must be positive numbers. Due to this condition, we can make the following substitution $atan2(x,y-1) = atan(\frac{y-1}{x})$:
$$0 \lt atan(\frac{y-1}{x}) \lt \frac{\pi}{4}$$
Apply tangent and simplify:
$$tan(0) \lt tan(atan(\frac{y-1}{x})) \lt tan(\frac{\pi}{4})$$
$$0 \lt \frac{y-1}{x} \lt 1$$
Multiply by $x$:
$$0 \lt y-1 \lt x$$
Add $+1$:
$$1 \lt y \lt x+1$$


So basically your target are all the complex numbers $z=x+yi$ whose position is below the line defined by $y=x+1$, in red color, not including the complex numbers belonging to the red line, and above the blue line $y=1$, not including the complex numbers belonging to the blue line itself.

