I was getting ready to type up a nice geometric solution to the second part that I found in Heath's translation of the Elements and I saw that you solved it. I think it's still interesting enough to add here, although I'll leave the details to anyone interested. It shows that the locus of a point such that its distances from two given points are in a given ratio (not equal to one) is a circle. In your case, you want all points z whose distances from 1 and -1 are in the ratio k

The following proposition can be extended to the case where the external angle EAC is bisected and meets line BC produced.

http://aleph0.clarku.edu/~djoyce/java/elements/bookVI/propVI3.html

Now if B and C are the two given points (1 and -1 in this case), there are two points X and Y on line BC that satisfy what we're after, one on segment BC and one outside segment BC. Construct the circle with XY as diameter and the extended proposition above can be used to prove that you have precisely the points in the complex plane that satisfy the condition of your second problem.