38104

Set:
z1=a+bi
z2=c+di

z3=z1*z2=(a+bi)(c+di)=(ac-bd)+(ad+bc)i

As z3 lies on the x axis Im(z3)=0 and let the given point on the x axis be m (i.e. Re(z3)=m)

ac-bd=m
ad+bc=0

Rearrange the second equation to obtain c:
ad=-bc
-(ad/b)=c

Substitute this into ac-bd=m:

-(a^2 d/b)-bd=m
d(-(a^2/b)-b)=m
m/(-(a^2/b)-b)=d

Multiply top and bottom by -b:
-(mb/(a^2+b^2))=d

Substitute this into -(ad/b)=c to get:
c=ma/(a^2+b^2)

So for z3=z1*z2=m and z1=a+bi
z2=ma/(a^2+b^2)-(mb/(a^2+b^2))i
=(m/(a^2+b^2))(a-bi)

So z2 will always be a scaler multiplied by a-bi. This means it will have he same trajectory as z1 reflected about the x-axis.