Where can I get a good text to read up on Complex variables I'm doing a study on calculus of complex variables. I started out by reading the topological concepts, curves and region in the complex plane. At the end exercises were given of which I did not know how to solve the following and they fall under loci problems.


*

*$ |\arg z| < \frac \pi 2 $

*$ - \pi < \Im(z) < \pi $

*$ |z-1| + |z+1| = 3 $

Here's my attempt

$ | z - 1 | + | z + 1 | = 3 $
$ | z + 1 | = 3 - | z + 1 | $
Squaring both sides
$ | z + 1 | ^ 2 = 9 - 6 | z - 1 | + | z - 1 | ^ 2 $
Simplifying
$ 4 x = 9 - 6 | z - 1 | $
Squaring again and simplifying
$ - 10 x ^ 2 + 6 y ^ 2 + 60 x = 75 $
and quite a few others.

Please I need a guide as the text I'm studying with did not provide sufficient information for me to be able to solve these problems. If I can get a reference material to read up on to solidify my knowledge in this area I'll be grateful. Also if I can get the solution to the problems I'll also appreciate that. Thanks.
 A: $1)$ As mentioned by @imranfat in the comments, the argument of $z$ is the angle between the positive real axis and the line joining the point to the origin, so since you want $-\frac{\pi}{2}<arg(z)<\frac{\pi}{2}$, these are covered in the $1^{st}$ and $4^{th}$ quadrants.
$2)$ Setting $z=x+iy$, it is clear that you want $-\pi <y<\pi$.
$3)$ Again we set $z=x+iy$ so that $|z|^{2}=x^{2}+y^{2}$ and we perform some algebraic manipulation. Squaring both sides gives:
$$9=(|z+1|+|z-1|)^{2}=|z+1|^{2}+|z-1|^{2}+2|z-1||z+1|=(x+1)^{2}+y^{2}+(x-1)^{2}+y^{2}+2|(x+iy)^{2}-1|=2x^{2}+2y^{2}+2+2\sqrt{(x^{2}-y^{2}-1)^{2}+(2xy)^{2}}$$ and now we divide both sides by 2 and re-arrange:
$$\frac{9}{2}-x^{2}-y^{2}-1=\sqrt{(x^{2}-y^{2}-1)^{2}+(2xy)^{2}}$$ and finally we square both sides and collect like terms:
$$0=(x^{2}-y^{2}-1)^{2}+(2xy)^{2}-(x^{2}+y^{2}-\frac{7}{2})^{2}=\frac{1}{4}(20x^{2}+36y^{2}-45).$$
Thus the required answer is all points which satisfy $20x^{2}+36^{2}=45$, which is an ellipse, centred at the origin with foci at $(-1,0)$ and $(1,0)$.
