If the point P represents the complex number | vedclass

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

Question

(B) Given $z=x+iy$, where $P=(x, y)$.Consider the expression $\frac{z+i}{z-1} = \frac{x+i(y+1)}{(x-1)+iy}$.To simplify, multiply the numerator and den

Vedclass · Accepted Answer

$x^2+y^2-x+y=0$ and $(x, y) 
eq (1,0)$

Vedclass · Answer

(B) Given $z=x+iy$, where $P=(x, y)$.Consider the expression $\frac{z+i}{z-1} = \frac{x+i(y+1)}{(x-1)+iy}$.To simplify, multiply the numerator and denominator by the conjugate of the denominator, $(x-1)-iy$:$\frac{x+i(y+1)}{(x-1)+iy} 	imes \frac{(x-1)-iy}{(x-1)-iy} = \frac{x(x-1) - ixy + i(y+1)(x-1) + y(y+1)}{(x-1)^2+y^2}$.Expanding the numerator:$= \frac{x^2-x + y^2+y + i(xy-x+y+1-xy)}{(x-1)^2+y^2} = \frac{(x^2+y^2-x+y) + i(1-x+y)}{(x-1)^2+y^2}$.Since $\frac{z+i}{z-1}$ is purely imaginary, its real part must be zero:$\operatorname{Re}\left(\frac{z+i}{z-1}ight) = 0 \Rightarrow \frac{x^2+y^2-x+y}{(x-1)^2+y^2} = 0$.This implies $x^2+y^2-x+y=0$, provided the denominator $(x-1)^2+y^2 
eq 0$, which means $(x, y) 
eq (1, 0)$.

If the point $P$ represents the complex number $z=x+iy$ in the Argand plane and if $\frac{z+i}{z-1}$ is a purely imaginary number, then the locus of $P$ is:

Similar Questions

Let $z = x + iy$ be a point in the Argand plane. If the amplitude of $\left(\frac{z - 3}{z + 2i}\right)$ is $\frac{\pi}{2}$,then the locus of $z$ is

Let $A = \{z \in \mathbb{C} : |z - 2| \le 4\}$ and $B = \{z \in \mathbb{C} : |z - 2| + |z + 2| = 5\}$. Then the maximum value of $\{|z_1 - z_2| : z_1 \in A \text{ and } z_2 \in B\}$ is

Let $C$ denote the set of all complex numbers. Define $A = \{(z, w) \mid z, w \in C \text{ and } |z| = |w|\}$ and $B = \{(z, w) \mid z, w \in C \text{ and } z^2 = w^2\}$. Then:

Let $S = \{z \in \mathbb{C} : |z-1|=1 \text{ and } (\sqrt{2}-1)(z+\bar{z}) - i(z-\bar{z}) = 2\sqrt{2}\}$. Let $z_1, z_2 \in S$ be such that $|z_1| = \max_{z \in S} |z|$ and $|z_2| = \min_{z \in S} |z|$. Then $|\sqrt{2}z_1 - z_2|^2$ equals:

The points $1 + 3i$,$5 + i$,and $3 + 2i$ in the complex plane are

Vedclass Test Series

Exam Paper Generator

Online Exam Module