For the first part a) it is very clear a circle with center at -i. With common sense using equation of a circle.
For the 2nd part i'm having some trouble. My steps..
b) x+2y<3
$$(\sqrt{x})^2 + (\sqrt{2y})^2 < (\sqrt{3})^2$$ which i guess is an ellipse?
can you always solve algebraicly and plot for the polynom? but sometimes it doesn't work. i always substitute z=a+ib and get an equation of a straight line inequality instead of a circle. How to find a reliable way to do this problem?
You might want to check your reasoning on (a) - $|z+i|<1$ is a circle centered at $-i$ - but we have $>$ instead of $<$, so we actually have that it's the whole plane minus a circle.
Also, as for $b$, I don't think your step of changing the absolute values to $\sqrt{x}^2$ is much use - and I think it's mislead you. Yes, you're right that that sum looks like the form for an ellipse, but it's in terms of $\sqrt{x}$ and $\sqrt{y}$, not in terms of $x$ and $y$, like it would have to be to really be an ellipse.
What would be better is to consider the four quadrants individually - e.g. in the quadrant where $\text{Re}(z)$ and $\text{Im}(z)$ are positive, we have that $$x+2y<3$$ which is conveniently just a little triangular slice with vertices at $0$, $3$ and $\frac{3}2i$. You can repeat this for every quadrant, negating $\text{Re}(z)$ and $\text{Im}(z)$ as necessary on each quadrant for the absolute value.