P is a point denoting z in the Argand diagram | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $z = x + iy$. Then $\frac{z-i}{z-1} = \frac{x + i(y-1)}{(x-1) + iy}$.To make this purely imaginary,we multiply by the conjugate of the denomin

Vedclass · Accepted Answer

the points on the circle with centre $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$,excluding the points $(1, 0)$ and $(0, 1)$

Vedclass · Answer

(C) Let $z = x + iy$. Then $\frac{z-i}{z-1} = \frac{x + i(y-1)}{(x-1) + iy}$.To make this purely imaginary,we multiply by the conjugate of the denominator: $\frac{(x + i(y-1))((x-1) - iy)}{(x-1)^2 + y^2}$.The real part is $\frac{x(x-1) + y(y-1)}{(x-1)^2 + y^2} = 0$.This implies $x^2 - x + y^2 - y = 0$,which is $x^2 + y^2 - x - y = 0$.Completing the square,we get $\left(x - \frac{1}{2}\right)^2 + \left(y - \frac{1}{2}\right)^2 = \left(\frac{1}{\sqrt{2}}\right)^2$.This represents a circle with centre $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$.Since the expression is undefined at $z = 1$ (i.e.,$(1, 0)$) and the numerator is zero at $z = i$ (i.e.,$(0, 1)$),these points must be excluded.

$P$ is a point denoting $z$ in the Argand diagram. If $\frac{z-i}{z-1}$ is always purely imaginary,then the locus of $P$ is

Similar Questions

For any real number $r$, let $A_r = \{e^{i \pi r n} : n \in \mathbb{N}\}$ be a set of complex numbers. Then,

If $z$ and $\omega$ are two non-zero complex numbers such that $|z \omega|=1$ and $\operatorname{Arg}(z) - \operatorname{Arg}(\omega) = \frac{\pi}{2}$,then $\bar{z} \omega =$

Let $z_1, z_2 \in \mathbb{C}$ such that $|z_1 + z_2| = \sqrt{3}$ and $|z_1| = |z_2| = 1$. Then the value of $|z_1 - z_2|$ is

In the complex plane $\mathbb{C}$,the set $\{z \in \mathbb{C} : \arg \left(\frac{z-1}{z+1}\right) = \frac{\pi}{4}\}$ represents

Let $S = \{z \in \mathbb{C} : |z-3| \leq 1 \text{ and } z(4+3i) + \bar{z}(4-3i) \leq 24\}$. If $\alpha + i\beta$ is the point in $S$ which is closest to $4i$,then $25(\alpha + \beta)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module