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Specifically b) and c)

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Confused with iii). Any help would be appreciated!

Did they make a mistake on line two, with x(x-a)?? [When I multiplied out, i got x(x+a), ending up with the locus as a circle with centre (-a/2, b/2) instead of (a/2, b/2)] also, is there supposed to be a restriction on the locus? like excluding (a,0) (because z =/ a on the denominator) Thanks!

In the answer to the question above on line 5 they seem to substitute an abs term inside of an abs term itself. A similar thing occurs on line 2, ie given that |x| = 1, |x - 2| can be written as |1 - 2|. However I thought that x = 1, -1, and so it can be written as either that or |-1-2|. Am I...

z1, z2 and z3 lie on a circle. The origin also lies on the circle Prove 1/z1, 1/z2 and 1/z3 are collinear