Why do we flip the inequality sign when removing the log?

Because the base $\frac{1}{\sqrt{2}}$ is less than $1$. In logarithmic functions with a base between $0$ and $1$, larger inputs yield smaller outputs. Therefore, to maintain the inequality, we must reverse the direction of the comparison.

Why is the condition $|z| \neq 1$ given?

This is a domain constraint. If $|z| = 1$, the denominator $(|z|-1)^2$ becomes zero, making the expression undefined. We must exclude this point from our final solution set.

Why do we ignore the negative root when solving the quadratic?

The variable $|z|$ represents the modulus of a complex number, which is by definition a distance from the origin. Distances cannot be negative, so we must restrict our solution to $|z| \geq 0$.

JEE Main 2021 (16 March Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: Let a complex number $z, ∣ z ∣ \neq = 1$ , satisfy $lo g_{\frac{1}{2}} (\frac{∣ z ∣ + 11}{( ∣ z ∣ - 1 ) ^{2}}) \leq 2$ . Then, the largest value of $∣ z ∣$ is equal to

Select Answer:

Progressive Hints

Hint 1

The base of the logarithm is

\frac{1}{2}

, which is approximately

0.707

. Since this base is strictly between

0

and

1

, the logarithmic function is monotonically decreasing. This is the crucial trap: when you remove the logarithm, the inequality sign must flip.

Hint 2

Recall the fundamental property of logarithmic inequalities: for a base

b

where

0 < b < 1

, the inequality

lo g_{b} (x) \leq y

is equivalent to

x \geq b^{y}

. Apply this to your expression to transform the logarithmic inequality into a rational algebraic inequality.

Hint 3

After clearing the denominator by multiplying both sides by

(∣ z ∣ - 1)^{2}

(which is always positive since

∣ z ∣ \neq = 1

), you will arrive at a quadratic inequality in terms of

∣ z ∣

. Solve

∣ z ∣^{2} - 4∣ z ∣ - 21 \leq 0

to find the valid range for the magnitude of

z

Visualized Solution

Analyze the Logarithmic Inequality

Given inequality: $lo g_{\frac{1}{2}} (\frac{∣ z ∣ + 11}{( ∣ z ∣ - 1 ) ^{2}}) \leq 2$
Initial condition: $∣ z ∣ \neq = 1$ (to ensure the denominator is non-zero).

Evaluate the Logarithm Base

Base of the logarithm: $b = \frac{1}{2} \approx 0.707$
Since $0 < b < 1$ , the function $f (x) = lo g_{b} (x)$ is a decreasing function.
Property: $lo g_{b} (x) \leq y ⟹ x \geq b^{y}$ when $0 < b < 1$ .

Remove the Logarithm

Removing the log and reversing the inequality sign:
$\frac{∣ z ∣ + 11}{( ∣ z ∣ - 1 ) ^{2}} \geq (\frac{1}{2})^{2}$

Simplify the Right Hand Side

Simplify the power on the right hand side:
$\frac{∣ z ∣ + 11}{( ∣ z ∣ - 1 ) ^{2}} \geq \frac{1}{2}$

Cross-Multiply to Clear Fractions

Cross-multiplying (since $(∣ z ∣ - 1)^{2} > 0$ ):
$2 (∣ z ∣ + 11) \geq (∣ z ∣ - 1)^{2}$

Expand the Squared Term

Expand the brackets and the perfect square:
$2∣ z ∣ + 22 \geq ∣ z ∣^{2} - 2∣ z ∣ + 1$

Rearrange into Quadratic Form

Rearrange all terms to one side to form a quadratic:
$0 \geq ∣ z ∣^{2} - 4∣ z ∣ - 21$
$⟹ ∣ z ∣^{2} - 4∣ z ∣ - 21 \leq 0$

Factorize the Quadratic Expression

Factorizing the quadratic expression:
Find two numbers that multiply to $- 21$ and add to $- 4$ : $- 7$ and $3$ .
$(∣ z ∣ - 7) (∣ z ∣ + 3) \leq 0$

Determine the Range of $∣ z ∣$

Solving the inequality gives $- 3 \leq ∣ z ∣ \leq 7$ .
Since the modulus $∣ z ∣$ must be non-negative: $0 \leq ∣ z ∣ \leq 7$ .
Also, we were given $∣ z ∣ \neq = 1$ , so the actual set is $[0, 1) \cup (1, 7]$ .

Identify the Largest Value

The range for $∣ z ∣$ is $[0, 7]$ excluding $1$ .
The largest value in this range is $7$ .
Key Takeaway: When the base of a logarithm is less than 1, the inequality sign reverses.

00:00 / 00:00

The Sigma Insight: Conjugate and Modulus

Solution Diagram

Analyzing the Logarithmic Constraint

We are given the inequality:

lo g_{\frac{1}{2}} (\frac{∣ z ∣ + 11}{( ∣ z ∣ - 1 ) ^{2}}) \leq 2

The base of the logarithm is

b = \frac{1}{2}

. Since

0 < b < 1

, the logarithmic function is strictly decreasing. Therefore, when we remove the logarithm, we must reverse the inequality sign.

Transforming the Inequality

Applying the property of logarithms, we transform the expression into:

\frac{∣ z ∣ + 11}{( ∣ z ∣ - 1 ) ^{2}} \geq (\frac{1}{2})^{2}

Simplifying the right side, we obtain:

\frac{∣ z ∣ + 11}{( ∣ z ∣ - 1 ) ^{2}} \geq \frac{1}{2}

Since

(∣ z ∣ - 1)^{2} > 0

for $|z| eq 1$, we can safely cross-multiply without changing the direction of the inequality:

2 (∣ z ∣ + 11) \geq (∣ z ∣ - 1)^{2}

Solving the Quadratic Siege

Expanding the terms on both sides, we get:

2∣ z ∣ + 22 \geq ∣ z ∣^{2} - 2∣ z ∣ + 1

Rearranging all terms to one side to form a standard quadratic inequality:

∣ z ∣^{2} - 4∣ z ∣ - 21 \leq 0

We factorize the quadratic expression by finding two numbers that multiply to

- 21

and add to

- 4

, which are

- 7

and

3

(∣ z ∣ - 7) (∣ z ∣ + 3) \leq 0

Determining the Final Range

The roots of the quadratic are

∣ z ∣ = 7

and

∣ z ∣ = - 3

. The inequality holds between these roots:

- 3 \leq ∣ z ∣ \leq 7

Because

∣ z ∣

represents a modulus (distance), it must be non-negative (

∣ z ∣ \geq 0

). Additionally, the original expression is undefined at

∣ z ∣ = 1

, so we must exclude this point.

The valid range for

∣ z ∣

[0, 1) \cup (1, 7]

. Consequently, the largest possible value of

∣ z ∣

is $7$ .

Animated Solution for Mathematics - Complex Numbers: Let a complex number $z, ∣ z ∣ \neq = 1$ , satisfy $lo g_{\frac{1}{2}} (\frac{∣ z ∣ + 11}{( ∣ z ∣ - 1 ) ^{2}}) \leq 2$ . Then, the largest value of $∣ z ∣$ is equal to

Select Answer:

Visualized Solution

Analyze the Logarithmic Inequality

Evaluate the Logarithm Base

Remove the Logarithm

Simplify the Right Hand Side

Cross-Multiply to Clear Fractions

Expand the Squared Term

Rearrange into Quadratic Form

Factorize the Quadratic Expression

Determine the Range of $∣ z ∣$

Identify the Largest Value

The Sigma Insight: Conjugate and Modulus

Analyzing the Logarithmic Constraint

Transforming the Inequality

Solving the Quadratic Siege

Determining the Final Range

Similar Questions

If $z \neq = 0$ be a complex number such that $∣ z - \frac{1}{z} ∣ = 2$ , then the maximum value of $∣ z ∣$ is:

The least value of $∣ z ∣$ where $z$ is complex number which satisfies the inequality $exp (\frac{( ∣ z ∣ + 3 ) ( ∣ z ∣ - 1 )}{∣∣ z ∣ + 1∣} lo g_{e} 2) \geq lo g_{2} ∣5 7 + 9 i ∣$ , $i = - 1$ , is equal to :

If $∣ z - 4/ z ∣ = 2$ , then the maximum value of $∣ Z ∣$ is equal to

Let $z$ and $ω$ be two complex numbers such that $∣ z ∣ \leq 1, ∣ ω ∣ \leq 1$ and $∣ z + iω ∣ = ∣ z - i \overset{ω}{ˉ} ∣ = 2$ then $z$ equals

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If $z_{1}, z_{2}$ and $z_{3}$ are complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ = ∣ z_{3} ∣ = ∣1/ z_{1} + 1/ z_{2} + 1/ z_{3} ∣ = 1$ , then $∣ z_{1} + z_{2} + z_{3} ∣$ is

If $z = x + i y, x y \neq = 0$ , satisfies the equation $z^{2} + i \overset{z}{ˉ} = 0$ , then $∣ z ∣^{2}$ is equal to :

Let $z$ be complex number satisfying $∣ z ∣^{3} + 2 z^{2} + 4 \overset{z}{ˉ} - 8 = 0$ , where $\overset{z}{ˉ}$ denotes the complex conjugate of $z$ . Let the imaginary part of $z$ be nonzero. Match each entry in List-I to the correct entries in List-II.

List-I

List-II

If $\frac{w - w ˉ z}{1 - z}$ is purely real where $w = α + i β, β \neq = 0$ and $z \neq = 1$ , then the set of the values of $z$ is

If $a > 0$ and $z = \frac{( 1 + i ) ^{2}}{a - i}$ , has magnitude $2/5$ , then $z$ is equal to :

Select Answer:

Visualized Solution

Analyze the Logarithmic Inequality

Evaluate the Logarithm Base

Remove the Logarithm

Simplify the Right Hand Side

Cross-Multiply to Clear Fractions

Expand the Squared Term

Rearrange into Quadratic Form

Factorize the Quadratic Expression

Determine the Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣z∣

Identify the Largest Value

The Sigma Insight: Conjugate and Modulus

Analyzing the Logarithmic Constraint

Transforming the Inequality

Solving the Quadratic Siege

Determining the Final Range

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z=0 be a complex number such that ∣z−z1​∣=2, then the maximum value of ∣z∣ is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If ∣z−4/z∣=2, then the maximum value of ∣Z∣ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z and ω be two complex numbers such that ∣z∣≤1,∣ω∣≤1 and ∣z+iω∣=∣z−iωˉ∣=2 then z equals

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​ and z2​ are two complex numbers such taht ∣z1​∣<1<∣z2​∣ then prove that ∣z1​−z2​1−z1​zˉ2​​∣<1.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​,z2​ and z3​ are complex numbers such that ∣z1​∣=∣z2​∣=∣z3​∣=∣1/z1​+1/z2​+1/z3​∣=1, then ∣z1​+z2​+z3​∣ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z=x+iy,xy=0, satisfies the equation z2+izˉ=0, then ∣z∣2 is equal to :

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 1−zw−wˉz​ is purely real where w=α+iβ,β=0 and z=1, then the set of the values of z is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a>0 and z=a−i(1+i)2​, has magnitude 2/5​, then z is equal to :

Determine the Range of $∣ z ∣$

If $z \neq = 0$ be a complex number such that $∣ z - \frac{1}{z} ∣ = 2$ , then the maximum value of $∣ z ∣$ is:

If $∣ z - 4/ z ∣ = 2$ , then the maximum value of $∣ Z ∣$ is equal to

Let $z$ and $ω$ be two complex numbers such that $∣ z ∣ \leq 1, ∣ ω ∣ \leq 1$ and $∣ z + iω ∣ = ∣ z - i \overset{ω}{ˉ} ∣ = 2$ then $z$ equals

If $z_{1}$ and $z_{2}$ are two complex numbers such taht $∣ z_{1} ∣ < 1 < ∣ z_{2} ∣$ then prove that $∣ \frac{1 - z _{1} z ˉ _{2}}{z _{1} - z _{2}} ∣ < 1$ .

If $z_{1}, z_{2}$ and $z_{3}$ are complex numbers such that $∣ z_{1} ∣ = ∣ z_{2} ∣ = ∣ z_{3} ∣ = ∣1/ z_{1} + 1/ z_{2} + 1/ z_{3} ∣ = 1$ , then $∣ z_{1} + z_{2} + z_{3} ∣$ is

If $z = x + i y, x y \neq = 0$ , satisfies the equation $z^{2} + i \overset{z}{ˉ} = 0$ , then $∣ z ∣^{2}$ is equal to :

If $\frac{w - w ˉ z}{1 - z}$ is purely real where $w = α + i β, β \neq = 0$ and $z \neq = 1$ , then the set of the values of $z$ is

If $a > 0$ and $z = \frac{( 1 + i ) ^{2}}{a - i}$ , has magnitude $2/5$ , then $z$ is equal to :