Question:

If $P$ is the affix of $z$ in the Argand diagram and $P$ moves so that $\frac{z-i}{z-1}$ is always purely imaginary, then locus of $z$ is

Updated On: Jul 2, 2022
  • circle, centre (2, 2), radius $\frac{1}{2}$
  • circle, centre $\left(-\frac{1}{2},-\frac{1}{2}\right)$, radius $\frac{1}{\sqrt2}$
  • circle, $\left(\frac{1}{2},\frac{1}{2}\right)$, radius $\frac{1}{\sqrt2}$
  • none of these
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The Correct Option is D

Solution and Explanation

$\frac{z-i}{z-1}=\frac{x+iy-1}{x+iy-1}$ $=\frac{x+i\left(y-1\right)}{x-1+iy} \cdot \frac{\left(x-1\right)-iy}{\left(x-1\right)-iy}$ $=\frac{x\left(x-1\right)+y\left(y-1\right)+i\left[\left(x-1\right)\left(y-1\right)-xy\right]}{\left(x-1\right)^{2}+y^{2}}$ Since $\frac{z-i}{z-1}$ is purely imaginary, $\therefore x^{2}+y^{2}-x-y=0$, which is a circle with centre $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $=\frac{1}{\sqrt{2}}$.
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Concepts Used:

Complex Numbers and Quadratic Equations

Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.