Loci of complex variables

Question

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.

1. |arg z| < π/2 
2. -π < im(z) < π
3. |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
 4x = 9 - 6|z-1| 
 Squaring again and simplifying
 -10x^2 + 6y^2 + 60x = 75

I have no idea on how to go about 1 and 2 specifically what gives me headache is the argz in the modulus it would have been an easy solve. Don't get me wrong I know what arg of z is all I'm saying is how does it help me get the loci that satifies that condition.

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

Please don't just give me the answers show me where you learnt it from.

Answer 1

|z-a| gives the distance from z to a, therefore |z-1| + |z+1| = 3 give the geometric loci where the sum of the distances to 1 and -1 is constant =3 this is the definition of an ellipse with focal points +1 and -1 and axes a=2.5, b=sqrt(3.25) so your calculation seems wrong

-pi/2<arg(z)<pi/2 ist the positive halfplane.

-pi<im(z)<pi is the strip between the two lines y=-pi/2 and y=+pi/2

Answer 2

There is an error in your solution. It should be $$\frac{4x^2}{9}+\frac{4y^2}{5}=1$$ Also, since $\displaystyle-\frac{π}{2}<\arg z<\frac{π}{2}$, you've to consider only half of the ellipse.

You don't have to worry about the second conditions as by default $\displaystyle-\frac{\sqrt 5}{2}<\text{Im}(z)<\frac{\sqrt 5}{2}$. The locus will look like this