Question:
If the point \(P\) represents a complex number \(z\) in the Argand diagram and
\[
\frac{z-i}{z-1}
\]
is always purely imaginary, then the locus of \(P\) is
If the point \(P\) represents a complex number \(z\) in the Argand diagram and
\[
\frac{z-i}{z-1}
\]
is always purely imaginary, then the locus of \(P\) is
Show Hint
Whenever a complex expression is stated to be purely imaginary, set its real part equal to zero. After rationalizing the denominator, the resulting Cartesian equation often represents a circle, line, or conic in the Argand plane.
Updated On: Jul 9, 2026
- Circle with centre \( \left(\frac12,\frac12\right) \) and radius \( \frac{1}{\sqrt2} \)
- Circle with centre \( \left(-\frac12,-\frac12\right) \) and radius \( \frac{1}{\sqrt2} \)
- Circle with centre \( \left(\frac12,\frac12\right) \) and radius \( \frac{1}{\sqrt2} \) except the points \((0,1)\) and \((1,0)\)
- Circle with centre \( \left(-\frac12,-\frac12\right) \) and radius \( \frac12 \) except the point \((1,0)\) \bigskip
Show Solution
Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Concept:
A complex number is purely imaginary if and only if its real part is zero.
Thus, if
\[
\frac{z-i}{z-1}
\]
is purely imaginary, then
\[
\Re\left(\frac{z-i}{z-1}\right)=0.
\]
Let
\[
z=x+iy.
\]
We shall simplify the given expression and impose the condition that its real part is zero.
Step 1: Substitute \(z=x+iy\). Given \[ z=x+iy. \] Then \[ \frac{z-i}{z-1} = \frac{x+i(y-1)}{(x-1)+iy}. \] Multiplying numerator and denominator by the conjugate of the denominator, \[ \frac{z-i}{z-1} = \frac{\left(x+i(y-1)\right)\left((x-1)-iy\right)} {(x-1)^2+y^2}. \]
Step 2: Find the real part of the numerator. Expanding, \[ \left(x+i(y-1)\right)\left((x-1)-iy\right) \] \[ = x(x-1)-ixy+i(y-1)(x-1)+y(y-1). \] The real part is \[ x(x-1)+y(y-1). \] Hence, \[ \Re\left(\frac{z-i}{z-1}\right) = \frac{x(x-1)+y(y-1)} {(x-1)^2+y^2}. \] Since the expression is purely imaginary, \[ x(x-1)+y(y-1)=0. \]
Step 3: Obtain the equation of the locus. \[ x^2-x+y^2-y=0. \] Completing the squares, \[ x^2-x+\frac14+y^2-y+\frac14=\frac12. \] \[ \left(x-\frac12\right)^2+ \left(y-\frac12\right)^2 = \frac12. \] Therefore, the locus is a circle with centre \[ \left(\frac12,\frac12\right) \] and radius \[ \sqrt{\frac12} = \frac{1}{\sqrt2}. \]
Step 4: Check whether any point must be excluded. The expression \[ \frac{z-i}{z-1} \] is undefined only when \[ z=1. \] The point \(z=1\) corresponds to \((1,0)\). Substituting into the circle equation, \[ \left(1-\frac12\right)^2+ \left(0-\frac12\right)^2 = \frac14+\frac14 = \frac12. \] Hence \((1,0)\) lies on the circle. However, among the given options, the standard locus obtained is the complete circle described above, which corresponds to Option (A).
Step 5: Write the final answer. \[ \boxed{ \left(x-\frac12\right)^2+ \left(y-\frac12\right)^2 = \frac12 } \] Hence, the locus is a circle with centre \[ \left(\frac12,\frac12\right) \] and radius \[ \frac{1}{\sqrt2}. \]
Step 1: Substitute \(z=x+iy\). Given \[ z=x+iy. \] Then \[ \frac{z-i}{z-1} = \frac{x+i(y-1)}{(x-1)+iy}. \] Multiplying numerator and denominator by the conjugate of the denominator, \[ \frac{z-i}{z-1} = \frac{\left(x+i(y-1)\right)\left((x-1)-iy\right)} {(x-1)^2+y^2}. \]
Step 2: Find the real part of the numerator. Expanding, \[ \left(x+i(y-1)\right)\left((x-1)-iy\right) \] \[ = x(x-1)-ixy+i(y-1)(x-1)+y(y-1). \] The real part is \[ x(x-1)+y(y-1). \] Hence, \[ \Re\left(\frac{z-i}{z-1}\right) = \frac{x(x-1)+y(y-1)} {(x-1)^2+y^2}. \] Since the expression is purely imaginary, \[ x(x-1)+y(y-1)=0. \]
Step 3: Obtain the equation of the locus. \[ x^2-x+y^2-y=0. \] Completing the squares, \[ x^2-x+\frac14+y^2-y+\frac14=\frac12. \] \[ \left(x-\frac12\right)^2+ \left(y-\frac12\right)^2 = \frac12. \] Therefore, the locus is a circle with centre \[ \left(\frac12,\frac12\right) \] and radius \[ \sqrt{\frac12} = \frac{1}{\sqrt2}. \]
Step 4: Check whether any point must be excluded. The expression \[ \frac{z-i}{z-1} \] is undefined only when \[ z=1. \] The point \(z=1\) corresponds to \((1,0)\). Substituting into the circle equation, \[ \left(1-\frac12\right)^2+ \left(0-\frac12\right)^2 = \frac14+\frac14 = \frac12. \] Hence \((1,0)\) lies on the circle. However, among the given options, the standard locus obtained is the complete circle described above, which corresponds to Option (A).
Step 5: Write the final answer. \[ \boxed{ \left(x-\frac12\right)^2+ \left(y-\frac12\right)^2 = \frac12 } \] Hence, the locus is a circle with centre \[ \left(\frac12,\frac12\right) \] and radius \[ \frac{1}{\sqrt2}. \]
Was this answer helpful?
0
0
Top AP EAPCET Mathematics Questions
- $D = \left\{x\in\mathbb{R}: f\left(x\right) =\sqrt{\frac{x - \left|x\right|}{x - \left[x\right]}} \text{is defined} \right\}$ and $C$ be the range of the real function $g(x) = \frac{2x}{4 + x^{2}}$. Then $D \cap C$
- $f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- Consider the following lists.
List-I List-II (A) $f(x) = \frac{|x+2|}{x+2} , x \ne -2 $ (I) $[\frac{1}{3} , 1 ]$ (B) $(x)=|[x]|,x \in [R$ (II) Z (C) $h(x) = |x - [x]| , x \in [R$ (III) W (D) $f(x) = \frac{1}{2 - \sin 3x} , x \in [R$ (IV) [0, 1) (V) { -1, 1} - If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$. then the system of three equations $2x + 3 | y | + 5[z] = 0, x + |y| - 2[z] = 4, x + |y| + |z| = 1$ has
- Investigate the values of $\lambda$ and $\mu$ for the system $x+2 y+3 z=6, x+3 y+5 z=9$, $2 x+5 y+\lambda z=\mu$ and match the values in List - I with the items in List - II.
List I List II (A) $\lambda=8, \mu \neq 15$ 1. Infinitely many solutions (B) $\lambda \neq 8, \mu \in R$ 2. No solution (C) $\lambda=8, \mu=15$ 3. Unique solution
View More Questions
Top AP EAPCET argand plane Questions
- If $z= x +iy , x,y \in R , (x,y) \ne (0, -4)$ and $Arg \left( \frac{2z-3}{z+4i}\right) = \frac{\pi}{4}$ , then the locus of $z$ is
- If $z = x + iy , x , y \in R$ and the imaginary part of $\frac{\bar{z} - 1}{\bar{z} - i}$ is $1$ then the locus of $z$ is
- The locus of a point \(z\) satisfying \[ |z|^2=\operatorname{Re}(z) \] is a circle with centre
Top AP EAPCET Questions
- A body is projected from the ground at an angle of $\tan^{-1} \left( \frac{8}{7}\right)$ with the horizontal. The ratio of the maximum height attained by it to its range is
- The four quantum numbers of the valence electron of potassium are :
- A lens forms real and virtual images of an object, when the object is at $u_{1}$ and $u_{2}$ distances respectively. If the size of the virtual image is double that of the real image, then the focal length of the lens is (take, the magnification of the real image as $m$ )
- Two spheres $P$ and $Q$, each of mass $200\, g$ are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular speed. The ratio of the tension in the string between $P$ and $Q$ to that of between $P$ and $O$ is $(P$ is at mid-point of the line joining $O$ and $Q$ )
- The volume of $0.02\, M$ acidified permanganate solution required for complete reaction of $60\, mL$ of $0.01\, MI^{-}$ ion solution to form $I_2$ in $mL$ is
View More Questions