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Prove the following.
Prove that locus of z is circle and find its centre and radius if (z-i)/(z-1) is purely imaginary.
Prove that locus of $z$ is circle and find its centre and radius if $\frac{z-i}{z-1}$ is purely imaginary.
Given: $\frac{z-i}{z-1} \text{ is purely imaginary.}$
To show: $\text{Locus of } z \text{ is circle with centre } (1/2, 1/2) \text{ and radius } 1/\sqrt{2}$
Answer
Answer
Put $z = x + iy$ and make the real part of $\frac{z-i}{z-1}$ zero.
- Step 1: Let $z = x + iy$. Then $\frac{z-i}{z-1} = \frac{x + i(y-1)}{(x-1) + iy}$.
- Step 2: Multiply numerator and denominator by the conjugate $(x-1) - iy$: $\frac{[x + i(y-1)][(x-1) - iy]}{(x-1)^2 + y^2}$.
- Step 3: Real part of the numerator $= x(x-1) + y(y-1)$ (using $i^2 = -1$), imaginary part $= (y-1)(x-1) - xy = 1 - x - y$.
- Step 4: The number is purely imaginary, so its real part is zero: $x(x-1) + y(y-1) = 0$, with $z \neq 1$ and $z \neq i$.
- Step 5: $x^2 + y^2 - x - y = 0$.
- Step 6: Completing the squares: $\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.
- Step 7: This is the equation of a circle, so the locus of $z$ is a circle with centre $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$.