The region represented by {z=x+iy in mathbb{C | vedclass

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

Question

(D) Given the inequality $|z|-\operatorname{Re}(z) \leq 1$,where $z = x+iy$.Substituting the values,we get $\sqrt{x^{2}+y^{2}} - x \leq 1$.Rearranging

Vedclass · Accepted Answer

$y^{2} \leq 2\left(x+\frac{1}{2}\right)$

Vedclass · Answer

(D) Given the inequality $|z|-\operatorname{Re}(z) \leq 1$,where $z = x+iy$.Substituting the values,we get $\sqrt{x^{2}+y^{2}} - x \leq 1$.Rearranging the terms,we have $\sqrt{x^{2}+y^{2}} \leq 1+x$.Since the square root is non-negative,$1+x$ must be $\geq 0$,i.e.,$x \geq -1$.Squaring both sides,we get $x^{2}+y^{2} \leq (1+x)^{2}$.Expanding the right side,$x^{2}+y^{2} \leq 1+2x+x^{2}$.Subtracting $x^{2}$ from both sides,we get $y^{2} \leq 2x+1$.Factoring out $2$,we obtain $y^{2} \leq 2\left(x+\frac{1}{2}\right)$.

The region represented by $\{z=x+iy \in \mathbb{C} : |z|-\operatorname{Re}(z) \leq 1\}$ is also given by the inequality

Similar Questions

Let $|z_1 - 8 - 2i| \leq 1$ and $|z_2 - 2 + 6i| \leq 2$,where $z_1, z_2 \in \mathbb{C}$. Then the minimum value of $|z_1 - z_2|$ is:

If $z_1, z_2, z_3$ are vertices of a triangle in the Argand plane such that $|z_1 - z_2| = |z_1 - z_3|$,then $\arg \left( \frac{2z_1 - z_2 - z_3}{z_3 - z_2} \right)$ is:

For any complex number $z$, the minimum value of $|z| + |z - 1|$ is

$A$ complex number $z$ is such that $arg\left( \frac{z - 2}{z + 2} \right) = \frac{\pi}{3}$. The points representing this complex number will lie on

If the complex numbers $z_1, z_2, 0$ are vertices of an equilateral triangle,then $z_1^2 + z_2^2 =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module