I have hard time to try to figure out the locus of those points which satisfies these equations. Can someone give me some hints to find the locus? Thanks so much. I really appreciate.
Find the locus of $z$ which satisfies these equations in complex plane:
a) Im($z-z_1 \over z - z_2$) = $0$ ($z_1 \neq z_2$)
b) Re($z-z_1 \over z - z_2$) = $0$ ($z_1 \neq z_2$)
On
a) $\dfrac{z-z_1}{z-z_2}=\lambda i$, then $z=\dfrac{z_1-\lambda iz_2}{1-\lambda i}=z_2+\dfrac{z_1-z_2}{1-\lambda i}$. This is a similarity transform applied to the inversion of the line $1-\lambda i$, i.e. a circle (through the diameter $z_1z_2)$.
b) $\dfrac{z-z_1}{z-z_2}=\lambda$, then $z=\dfrac{z_1-\lambda z_2}{1-\lambda}$. This is a linear combination of $z_1$ and $z_2$, hence a straight line through them.
In the case (b), where the answer is purely imaginary, then the angle between zz1 and zz2 is a right angle. This means that z lies on the semicircle with z1 and z2 as the endpoints of the diameter.
In case (a), when the imaginary part is zero, we let the value be a non-zero r, where r is real. In this case we get z1 = rz2 + (1-r)z, which means z lies on the other side of the line joining z2 and z1 (beyond z1).