If z is a complex number such that frac{z-i}{ | vedclass

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

Question

(D) Given that $\frac{z-i}{z-1}$ is purely imaginary,the real part of $\frac{z-i}{z-1}$ must be $0$.Let $z = x+iy$. Then $\frac{z-i}{z-1} = \frac{x+i(

Vedclass · Accepted Answer

$2\sqrt{2}$

Vedclass · Answer

(D) Given that $\frac{z-i}{z-1}$ is purely imaginary,the real part of $\frac{z-i}{z-1}$ must be $0$.Let $z = x+iy$. Then $\frac{z-i}{z-1} = \frac{x+i(y-1)}{(x-1)+iy}$.Multiplying the numerator and denominator by the conjugate of the denominator $(x-1)-iy$,we get:$\frac{(x+i(y-1))((x-1)-iy)}{(x-1)^2+y^2} = \frac{x(x-1)+y(y-1) + i((y-1)(x-1)-xy)}{(x-1)^2+y^2}$.For this to be purely imaginary,the real part must be zero:$x(x-1)+y(y-1) = 0 \Rightarrow x^2-x+y^2-y = 0$.This represents a circle with center $C(\frac{1}{2}, \frac{1}{2})$ and radius $r = \sqrt{(\frac{1}{2})^2 + (\frac{1}{2})^2} = \frac{1}{\sqrt{2}}$.We want to find the minimum value of $|z-(3+3i)|$,which is the distance from point $P(3,3)$ to the circle.The distance from $P(3,3)$ to the center $C(\frac{1}{2}, \frac{1}{2})$ is $PC = \sqrt{(3-\frac{1}{2})^2 + (3-\frac{1}{2})^2} = \sqrt{(\frac{5}{2})^2 + (\frac{5}{2})^2} = \frac{5}{\sqrt{2}}$.The minimum distance is $PC - r = \frac{5}{\sqrt{2}} - \frac{1}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.

If $z$ is a complex number such that $\frac{z-i}{z-1}$ is purely imaginary,then the minimum value of $|z-(3+3i)|$ is:

Similar Questions

If $A = \{z = x + iy : \text{real part of } \frac{\bar{z}-1}{z-i} = 2\}$,then the locus of the point $P(x, y)$ in the Cartesian plane is:

If $\operatorname{Re}\left(\frac{z-1}{2z+i}\right)=1,$ where $z=x+iy,$ then the point $(x, y)$ lies on a

If $|Z_1|=|Z_2|=|Z_3|=1$ and $Z_1+Z_2+Z_3=0$, then the area of the triangle whose vertices are $Z_1, Z_2, Z_3$ is

The equation $|z - i| = |z - 1|$,where $i = \sqrt{-1}$,represents:

Let $S = \{z \in \mathbb{C} : z^{2} + \bar{z} = 0\}$. Then $\sum_{z \in S} (\operatorname{Re}(z) + \operatorname{Im}(z))$ is equal to $......$

Vedclass Test Series

Exam Paper Generator

Online Exam Module