Question

find the locus of z in re(z-i/z-2)=0

Answer 1

Answer:

You need to "element chase"; that is, suppose (x,y)∈A×(B∩C)(x,y)∈A×(B∩C) and then show you must have (x,y)∈(A×B)∩(A×C)(x,y)∈(A×B)∩(A×C), then do the other direction as well. I'll do one direction for you and you can try the other yourself.

Suppose (x,y)∈A×(B∩C)(x,y)∈A×(B∩C). Then x∈Ax∈A and y∈B∩Cy∈B∩C, meaning y∈By∈B and y∈Cy∈C. Because y∈By∈B, (x,y)∈A×B(x,y)∈A×B, and because y∈Cy∈C, (x,y)∈A×C(x,y)∈A×C. Therefore (x,y)∈(A×B)∩(A×C)(x,y)∈(A×B)∩(A×C). This tells us A×(B∩C)⊆(A×B)∩(A×C)

Proof:

A×(B∪C)={(x,y):x∈A,y∈B∪C}={(x,y):x∈A,y∈B or x∈A,y∈C}={(x,y):(x,y)∈A×B or (x,y)∈A×C}=(A×B)∪(A×C)A×(B∪C)={(x,y):x∈A,y∈B∪C}={(x,y):x∈A,y∈B or x∈A,y∈C}={(x,y):(x,y)∈A×B or (x,y)∈A×C}=(A×B)∪(A×C)

Therefore, A×(B∪C)=(A×B)∪(A×C)A×(B∪C)=(A×B)∪(A×C)