The region represented by ${z = x + iy \in C : |z| – \text{Re}(z) \le 1}$ is also given by the inequality: Options: (A) $y^2 \le x + \frac{1}{2}$ (B) $y^2 \le 2 \left(x+\frac{1}{2}\right)$ (C) $y^2 \ge x + 1$ (D) $y^2 \ge 2(x + 1)$

### Analyzing the Setup To demystify the inequality $|z| - \text{Re}(z) \le 1$, we represent the complex number $z$ in its Cartesian form, $z = x + iy$. In this coordinate system, the modulus is defined as $|z| = \sqrt{x^2 + y^2}$ and the real part is $\text{Re}(z) = x$. Substituting these into the original inequality, we obtain: $$\sqrt{x^2 + y^2} - x \le 1$$ ### Isolating the Variables To simplify the expression, we isolate the square root term by moving $x$ to the right side of the inequality: $$\sqrt{x^2 + y^2} \le x + 1$$ Because the square root function $\sqrt{x^2 + y^2}$ is always non-negative, the right side of the inequality must also be non-negative. This imposes a critical domain constraint: $$x + 1 \ge 0 \implies x \ge -1$$ ### Deriving the Geometric Boundary Now, we square both sides of the inequality to eliminate the radical: $$x^2 + y^2 \le (x + 1)^2$$ Expanding the right side yields: $$x^2 + y^2 \le x^2 + 2x + 1$$ By subtracting $x^2$ from both sides, the expression simplifies significantly to: $$y^2 \le 2x + 1$$ ### Final Geometric Interpretation We can rewrite this result in the standard form of a parabola: $$y^2 \le 2\left(x + \frac{1}{2}\right)$$ This inequality describes the region **inside (or on) a parabola** that opens to the right, with its vertex located at $(-1/2, 0)$. Combined with our earlier constraint $x \ge -1$, we conclude that the solution set is the **interior region of the parabola $y^2 = 2x + 1$**.

JEE Main 2020 - 6 Sep (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: The region represented by $z = x + i y \in C : ∣ z ∣- Re (z) \leq 1$ is also given by the inequality:

Q: Wait, why do we need the constraint $x \ge -1$?

That is a fantastic catch! Remember that $\sqrt{x^2 + y^2}$ is always non-negative. If $x + 1$ were negative, we would be saying that a non-negative number is less than or equal to a negative number, which is impossible. By enforcing $x + 1 \ge 0$, we ensure that our squaring operation is mathematically valid.

Q: Why is the region 'inside' the parabola?

The inequality $\le$ tells us that the distance from the origin (the focus) is less than or equal to the distance from the line $x = -1$ (the directrix). By definition, the locus of points equidistant from a focus and a directrix is a parabola. If the distance to the focus is smaller, you are inside the curve!

Q: Can we use polar coordinates instead?

Absolutely! If you substitute $z = r(\cos\theta + i\sin\theta)$, you get $r - r\cos\theta \le 1$, which simplifies to $r \le \frac{1}{1 - \cos\theta}$. This is the polar equation of a parabola. It is a very elegant way to solve it, but the Cartesian method is often more intuitive for visualizing the region.

Select Answer:

Visualized Solution

The Complex Inequality

Given: $∣ z ∣ - Re (z) \leq 1$
Let $z = x + i y$

Modulus and Real Part

Modulus: $∣ z ∣ = x^{2} + y^{2}$
Real Part: $Re (z) = x$

Substituting the Values

Substitute into the inequality:
$x^{2} + y^{2} - x \leq 1$

Isolating the Square Root

Move $x$ to the right side:
$x^{2} + y^{2} \leq x + 1$

Domain Constraint

The output of a square root is always non-negative: $x^{2} + y^{2} \geq 0$
Therefore, the right side must also be non-negative: $x + 1 \geq 0 ⟹ x \geq - 1$

Squaring the Inequality

Square both sides:
$(x^{2} + y^{2})^{2} \leq (x + 1)^{2}$
$x^{2} + y^{2} \leq x^{2} + 2 x + 1$

Canceling Terms

Subtract $x^{2}$ from both sides:
$y^{2} \leq 2 x + 1$

The Standard Parabola Form

Factor out $2$ on the right side:
$y^{2} \leq 2 (x + \frac{1}{2})$
This matches Option (2).

Geometric Interpretation

The equation $y^{2} = 2 (x + \frac{1}{2})$ represents a parabola.
Vertex: $(- \frac{1}{2}, 0)$
Focus: $(0, 0)$

The Solution Region

The inequality $\leq$ indicates the region inside the parabola.
This is the locus of points whose distance from the origin (focus) is $\leq$ their distance from the line $x = - 1$ (directrix).

00:00 / 00:00

The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup

To demystify the inequality

∣ z ∣ - Re (z) \leq 1

, we represent the complex number

z

in its Cartesian form,

z = x + i y

In this coordinate system, the modulus is defined as

∣ z ∣ = x^{2} + y^{2}

and the real part is

Re (z) = x

. Substituting these into the original inequality, we obtain:

x^{2} + y^{2} - x \leq 1

Isolating the Variables

To simplify the expression, we isolate the square root term by moving

x

to the right side of the inequality:

x^{2} + y^{2} \leq x + 1

Because the square root function

x^{2} + y^{2}

is always non-negative, the right side of the inequality must also be non-negative. This imposes a critical domain constraint:

x + 1 \geq 0 ⟹ x \geq - 1

Deriving the Geometric Boundary

Now, we square both sides of the inequality to eliminate the radical:

x^{2} + y^{2} \leq (x + 1)^{2}

Expanding the right side yields:

x^{2} + y^{2} \leq x^{2} + 2 x + 1

By subtracting

x^{2}

from both sides, the expression simplifies significantly to:

y^{2} \leq 2 x + 1

Final Geometric Interpretation

We can rewrite this result in the standard form of a parabola:

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

This inequality describes the region inside (or on) a parabola that opens to the right, with its vertex located at

(- 1/2, 0)

Combined with our earlier constraint

x \geq - 1

, we conclude that the solution set is the interior region of the parabola $y^{2} = 2 x + 1$ .

Common Traps & FAQs

Wait, why do we need the constraint

x \geq - 1

That is a fantastic catch! Remember that

x^{2} + y^{2}

is always non-negative. If

x + 1

were negative, we would be saying that a non-negative number is less than or equal to a negative number, which is impossible. By enforcing

x + 1 \geq 0

, we ensure that our squaring operation is mathematically valid.

Why is the region 'inside' the parabola?

The inequality

\leq

tells us that the distance from the origin (the focus) is less than or equal to the distance from the line

x = - 1

(the directrix). By definition, the locus of points equidistant from a focus and a directrix is a parabola. If the distance to the focus is smaller, you are inside the curve!

Can we use polar coordinates instead?

Absolutely! If you substitute

z = r (cos θ + i sin θ)

, you get

r - r cos θ \leq 1

, which simplifies to

r \leq \frac{1}{1 - c o s θ}

. This is the polar equation of a parabola. It is a very elegant way to solve it, but the Cartesian method is often more intuitive for visualizing the region.

Animated Solution for Mathematics - Complex Numbers: The region represented by $z = x + i y \in C : ∣ z ∣- Re (z) \leq 1$ is also given by the inequality:

Select Answer:

Visualized Solution

The Complex Inequality

Modulus and Real Part

Substituting the Values

Isolating the Square Root

Domain Constraint

Squaring the Inequality

Canceling Terms

The Standard Parabola Form

Geometric Interpretation

The Solution Region

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Isolating the Variables

Deriving the Geometric Boundary

Final Geometric Interpretation

Similar Questions

The area (in sq. units) of the region $S = {z \in C : ∣ z - 1∣ \leq 2; (z + \overset{z}{ˉ}) + i (z - \overset{z}{ˉ}) \leq 2, Im (z) \geq 0}$ is

Let $S = {z = x + i y : ∣ z - 1 + i ∣ \geq ∣ z ∣, ∣ z ∣ < 2, ∣ z + i ∣ = ∣ z - 1∣}$ . Then the set of all values of $x$ , for which $w = 2 x + i y \in S$ for some $y \in R$ , is

Let $S = {z \in C : ∣ z - 1∣ = 1 and (2 - 1) (z + \overset{z}{ˉ}) - i (z - \overset{z}{ˉ}) = 22}$ . 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 $∣ 2 z_{1} - z_{2} ∣^{2}$ equals :

The locus of $z$ which lies in shaded region (excluding the boundaries) is best represented by

Let $a, b \in R$ and $a^{2} + b^{2} \neq = 0$ . Suppose $S = {z \in C : z = \frac{1}{a + ib t}, t \in R, t \neq = 0}$ , where $i = - 1$ . If $z = x + i y$ and $z \in S$ , then $(x, y)$ lies on

Let $S_{1} = {z \in C : ∣ z ∣ \leq 5}$ , $S_{2} = {z \in C : Im (\frac{z + 1 - 3 i}{1 - 3 i}) \geq 0}$ and $S_{3} = {z \in C : Re (z) \geq 0}$ . Then the area of the region $S_{1} \cap S_{2} \cap S_{3}$ is :

Let the curve $z (1 + i) + \overline{z} (1 - i) = 4, z \in C$ , divide the region $∣ z - 3∣ \leq 1$ into two parts of areas $α$ and $β$ . Then $∣ α - β ∣$ equals:

The inequality $∣ z - 4∣ < ∣ z - 2∣$ represents the region given by

Let $z$ be a complex number such that $∣ z + 2∣ = 1$ and $Im (\frac{z + 1}{z + 2}) = \frac{1}{5}$ . Then the value of $∣ Re (\overset{z}{ˉ} + 2) ∣$ is

If $z$ is a complex number such that $∣ z ∣ \geq 2$ , then the minimum value of $∣ z + 1/2∣$

Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The region represented by z=x+iy∈C:∣z∣–Re(z)≤1 is also given by the inequality:

Select Answer:

Visualized Solution

The Complex Inequality

Modulus and Real Part

Substituting the Values

Isolating the Square Root

Domain Constraint

Squaring the Inequality

Canceling Terms

The Standard Parabola Form

Geometric Interpretation

The Solution Region

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Isolating the Variables

Deriving the Geometric Boundary

Final Geometric Interpretation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area (in sq. units) of the region S={z∈C:∣z−1∣≤2;(z+zˉ)+i(z−zˉ)≤2,Im(z)≥0} is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={z=x+iy:∣z−1+i∣≥∣z∣,∣z∣<2,∣z+i∣=∣z−1∣}. Then the set of all values of x, for which w=2x+iy∈S for some y∈R, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of z which lies in shaded region (excluding the boundaries) is best represented by

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a,b∈R and a2+b2=0. Suppose S={z∈C:z=a+ibt1​,t∈R,t=0}, where i=−1​. If z=x+iy and z∈S, then (x,y) lies on

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S1​={z∈C:∣z∣≤5}, S2​={z∈C:Im(1−3​iz+1−3​i​)≥0} and S3​={z∈C:Re(z)≥0}. Then the area of the region S1​∩S2​∩S3​ is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the curve z(1+i)+z(1−i)=4,z∈C, divide the region ∣z−3∣≤1 into two parts of areas α and β. Then ∣α−β∣ equals:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The inequality ∣z−4∣<∣z−2∣ represents the region given by

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z be a complex number such that ∣z+2∣=1 and Im(z+2z+1​)=51​. Then the value of ∣Re(zˉ+2)∣ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z is a complex number such that ∣z∣≥2, then the minimum value of ∣z+1/2∣

Animated Solution for Mathematics - Complex Numbers: The region represented by $z = x + i y \in C : ∣ z ∣- Re (z) \leq 1$ is also given by the inequality:

The area (in sq. units) of the region $S = {z \in C : ∣ z - 1∣ \leq 2; (z + \overset{z}{ˉ}) + i (z - \overset{z}{ˉ}) \leq 2, Im (z) \geq 0}$ is

Let $S = {z = x + i y : ∣ z - 1 + i ∣ \geq ∣ z ∣, ∣ z ∣ < 2, ∣ z + i ∣ = ∣ z - 1∣}$ . Then the set of all values of $x$ , for which $w = 2 x + i y \in S$ for some $y \in R$ , is

The locus of $z$ which lies in shaded region (excluding the boundaries) is best represented by

Let $a, b \in R$ and $a^{2} + b^{2} \neq = 0$ . Suppose $S = {z \in C : z = \frac{1}{a + ib t}, t \in R, t \neq = 0}$ , where $i = - 1$ . If $z = x + i y$ and $z \in S$ , then $(x, y)$ lies on

Let $S_{1} = {z \in C : ∣ z ∣ \leq 5}$ , $S_{2} = {z \in C : Im (\frac{z + 1 - 3 i}{1 - 3 i}) \geq 0}$ and $S_{3} = {z \in C : Re (z) \geq 0}$ . Then the area of the region $S_{1} \cap S_{2} \cap S_{3}$ is :

Let the curve $z (1 + i) + \overline{z} (1 - i) = 4, z \in C$ , divide the region $∣ z - 3∣ \leq 1$ into two parts of areas $α$ and $β$ . Then $∣ α - β ∣$ equals:

The inequality $∣ z - 4∣ < ∣ z - 2∣$ represents the region given by

Let $z$ be a complex number such that $∣ z + 2∣ = 1$ and $Im (\frac{z + 1}{z + 2}) = \frac{1}{5}$ . Then the value of $∣ Re (\overset{z}{ˉ} + 2) ∣$ is

If $z$ is a complex number such that $∣ z ∣ \geq 2$ , then the minimum value of $∣ z + 1/2∣$