Wait, why can we assume $\alpha$ and $1/\bar{\alpha}$ are collinear with the origin?

Great observation! Remember that $\bar{\alpha} = |\alpha|^2 / \alpha$. Therefore, $1/\bar{\alpha} = \alpha / |\alpha|^2$. Since $|\alpha|^2$ is a positive real number, $1/\bar{\alpha}$ is just a positive scalar multiple of $\alpha$. In the complex plane, multiplying a vector by a positive scalar doesn't change its direction, only its magnitude. Thus, they must lie on the same ray starting from the origin.

JEE Advanced 2013

LEVELJEE Advanced

Animated Solution for Mathematics - Complex Numbers: Let complex numbers $α$ and $1/ \overset{α}{ˉ}$ lie on circles $(x - x_{0})^{2} + (y - y_{0})^{2} = r^{2}$ and $(x - x_{0})^{2} + (y - y_{0})^{2} = 4 r^{2}$ respectively. If $z_{0} = x_{0} + i y_{0}$ satisfies the equation $2∣ z_{0} ∣^{2} = r^{2} + 2$ , then $∣ α ∣ =$

Q: Is it possible for $r=0$ in this problem?

Excellent question! If $r=0$, the circles would collapse into a single point, $z_0$. However, the problem defines two distinct circles with radii $r$ and $2r$. If $r=0$, the radius of the outer circle would also be zero, which contradicts the geometric setup. Therefore, we can safely assume $r \neq 0$ and divide by $r^2$ during our simplification.

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the complex plane. You have two concentric circles centered at

z_{0}

. The inner circle has radius

r

, and the outer circle has radius

2 r

. The key is to realize that

α

and

1/ \overset{α}{ˉ}

are not just random points; they are related by a scaling factor involving

∣ α ∣^{2}

, which means they lie on the same ray from the origin.

Hint 2

Hint 2 (The Arsenal): Use the fundamental identity

∣ z ∣^{2} = z \overset{z}{ˉ}

to expand the circle equations

∣ α - z_{0} ∣^{2} = r^{2}

and

∣ \frac{1}{α ˉ} - z_{0} ∣^{2} = 4 r^{2}

. This will generate cross-terms involving

α \overset{z}{ˉ}_{0} + \overset{α}{ˉ} z_{0}

. Define this sum as a single variable

K

to simplify the algebra.

Hint 3

Hint 3 (The Execution): After substituting

K

and simplifying, you will arrive at an equation involving

∣ α ∣^{2}

∣ z_{0} ∣^{2}

, and

r^{2}

. Use the given condition

2∣ z_{0} ∣^{2} = r^{2} + 2

to substitute for

∣ z_{0} ∣^{2}

and solve for

∣ α ∣^{2}

Visualized Solution

Visualizing the Concentric Circles

Let the center of both circles be $z_{0} = x_{0} + i y_{0}$ .
The inner circle has radius $r$ and equation $∣ z - z_{0} ∣ = r$ .
The outer circle has radius $2 r$ and equation $∣ z - z_{0} ∣ = 2 r$ .
We are given that $α$ lies on the inner circle, and $\frac{1}{α ˉ}$ lies on the outer circle.

The Collinearity of $α$ and $\frac{1}{α ˉ}$

Recall that $\overset{α}{ˉ} = \frac{∣ α ∣ ^{2}}{α}$ .
Therefore, $\frac{1}{α ˉ} = \frac{α}{∣ α ∣ ^{2}}$ .
This implies that $\frac{1}{α ˉ}$ is a positive scalar multiple of $α$ .
Hence, the origin $O$ , $α$ , and $\frac{1}{α ˉ}$ are collinear and lie on the same ray!

Setting up the Circle Equations

Since $α$ lies on the inner circle: $∣ α - z_{0} ∣ = r$
Since $\frac{1}{α ˉ}$ lies on the outer circle: $∣ \frac{1}{α ˉ} - z_{0} ∣ = 2 r$
We will square both sides of these equations to eliminate the square roots.

Expanding $∣ α - z_{0} ∣^{2} = r^{2}$

Using the identity $∣ z ∣^{2} = z \overset{z}{ˉ}$ :
$(α - z_{0}) (\overset{α}{ˉ} - \overset{z}{ˉ}_{0}) = r^{2}$
Expanding the terms:
$α \overset{α}{ˉ} - α \overset{z}{ˉ}_{0} - \overset{α}{ˉ} z_{0} + z_{0} \overset{z}{ˉ}_{0} = r^{2}$
$∣ α ∣^{2} + ∣ z_{0} ∣^{2} - (α \overset{z}{ˉ}_{0} + \overset{α}{ˉ} z_{0}) = r^{2}$ --- (Equation 1)

Expanding $∣ \frac{1}{α ˉ} - z_{0} ∣^{2} = 4 r^{2}$

Using the same identity $∣ z ∣^{2} = z \overset{z}{ˉ}$ :
$(\frac{1}{α ˉ} - z_{0}) (\frac{1}{α} - \overset{z}{ˉ}_{0}) = 4 r^{2}$
Expanding the terms:
$\frac{1}{α α ˉ} - \frac{z ˉ _{0}}{α ˉ} - \frac{z _{0}}{α} + z_{0} \overset{z}{ˉ}_{0} = 4 r^{2}$
$\frac{1}{∣ α ∣ ^{2}} + ∣ z_{0} ∣^{2} - (\frac{α z ˉ _{0} + α ˉ z _{0}}{∣ α ∣ ^{2}}) = 4 r^{2}$ --- (Equation 2)

Introducing the Substitution Variable $K$

Notice the common cross-term in both equations: $α \overset{z}{ˉ}_{0} + \overset{α}{ˉ} z_{0}$ .
Let $K = α \overset{z}{ˉ}_{0} + \overset{α}{ˉ} z_{0}$ .
Equation 1 becomes: $∣ α ∣^{2} + ∣ z_{0} ∣^{2} - K = r^{2}$
Equation 2 becomes: $\frac{1}{∣ α ∣ ^{2}} + ∣ z_{0} ∣^{2} - \frac{K}{∣ α ∣ ^{2}} = 4 r^{2}$

Eliminating the Variable $K$

From Equation 1, express $K$ as:
$K = ∣ α ∣^{2} + ∣ z_{0} ∣^{2} - r^{2}$
Substitute this expression for $K$ into Equation 2:
$\frac{1}{∣ α ∣ ^{2}} + ∣ z_{0} ∣^{2} - \frac{∣ α ∣ ^{2} + ∣ z _{0} ∣ ^{2} - r ^{2}}{∣ α ∣ ^{2}} = 4 r^{2}$

Simplifying the Combined Equation

Multiply the entire equation by $∣ α ∣^{2}$ to clear the denominators:
$1 + ∣ z_{0} ∣^{2} ∣ α ∣^{2} - (∣ α ∣^{2} + ∣ z_{0} ∣^{2} - r^{2}) = 4 r^{2} ∣ α ∣^{2}$
Group terms containing $∣ z_{0} ∣^{2}$ and constant terms:
$∣ z_{0} ∣^{2} (∣ α ∣^{2} - 1) - (∣ α ∣^{2} - 1) + r^{2} = 4 r^{2} ∣ α ∣^{2}$
Factor out $(∣ α ∣^{2} - 1)$ :
$(∣ α ∣^{2} - 1) (∣ z_{0} ∣^{2} - 1) + r^{2} = 4 r^{2} ∣ α ∣^{2}$ --- (Equation 3)

Applying the Condition $2∣ z_{0} ∣^{2} = r^{2} + 2$

We are given: $2∣ z_{0} ∣^{2} = r^{2} + 2$
Divide by 2: $∣ z_{0} ∣^{2} = \frac{r ^{2}}{2} + 1$
Subtract 1 from both sides: $∣ z_{0} ∣^{2} - 1 = \frac{r ^{2}}{2}$
Substitute this into Equation 3:
$(∣ α ∣^{2} - 1) (\frac{r ^{2}}{2}) + r^{2} = 4 r^{2} ∣ α ∣^{2}$

Solving the Algebraic Equation

Divide the entire equation by $r^{2}$ (since $r \neq = 0$ ):
$\frac{∣ α ∣ ^{2} - 1}{2} + 1 = 4∣ α ∣^{2}$
Multiply by 2 to clear the fraction:
$∣ α ∣^{2} - 1 + 2 = 8∣ α ∣^{2}$
$1 = 7∣ α ∣^{2} ⟹ ∣ α ∣^{2} = \frac{1}{7}$
Since modulus must be positive: $∣ α ∣ = \frac{1}{7}$

Conclusion and Key Takeaways

The value of $∣ α ∣$ is $\frac{1}{7}$ .
This matches Option 3 (or index 2 in our options list).
Key Concept: Representing circles in the complex plane using $∣ z - z_{0} ∣ = R$ .
Algebraic Strategy: Using a substitution variable $K$ to simplify coupled complex equations.

00:00 / 00:00

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

Imagine you are standing on the complex plane, looking at two concentric circles centered at

z_{0} = x_{0} + i y_{0}

. The inner circle has a radius of

r

, and the outer circle has a radius of

2 r

We are given that

α

sits on the inner circle, and

1/ \overset{α}{ˉ}

sits on the outer circle. The relationship between

α

and its conjugate is defined by

\overset{α}{ˉ} = ∣ α ∣^{2} / α

This implies that

1/ \overset{α}{ˉ} = α /∣ α ∣^{2}

. Because

∣ α ∣^{2}

is a positive real number,

1/ \overset{α}{ˉ}

is simply a scaled version of

α

, meaning the origin,

α

, and

1/ \overset{α}{ˉ}

are collinear.

The Algebraic Dance

Since

α

is on the inner circle, we have

∣ α - z_{0} ∣ = r

, which squares to:

∣ α - z_{0} ∣^{2} = r^{2}

Similarly, for the outer circle,

∣1/ \overset{α}{ˉ} - z_{0} ∣ = 2 r

, which squares to:

∣1/ \overset{α}{ˉ} - z_{0} ∣^{2} = 4 r^{2}

Using the identity

∣ z ∣^{2} = z \overset{z}{ˉ}

, the first equation expands to:

∣ α ∣^{2} + ∣ z_{0} ∣^{2} - (α \overset{z}{ˉ}_{0} + \overset{α}{ˉ} z_{0}) = r^{2}

We define this as Equation 1. Expanding the second equation yields:

\frac{1}{∣ α ∣ ^{2}} + ∣ z_{0} ∣^{2} - \frac{α z ˉ _{0} + α ˉ z _{0}}{∣ α ∣ ^{2}} = 4 r^{2}

Let

K = α \overset{z}{ˉ}_{0} + \overset{α}{ˉ} z_{0}

. Our system simplifies to:

∣ α ∣^{2} + ∣ z_{0} ∣^{2} - K = r^{2}

\frac{1}{∣ α ∣ ^{2}} + ∣ z_{0} ∣^{2} - \frac{K}{∣ α ∣ ^{2}} = 4 r^{2}

The Final Reveal

To eliminate

K

, we substitute

K = ∣ α ∣^{2} + ∣ z_{0} ∣^{2} - r^{2}

into the second equation:

\frac{1}{∣ α ∣ ^{2}} + ∣ z_{0} ∣^{2} - \frac{∣ α ∣ ^{2} + ∣ z _{0} ∣ ^{2} - r ^{2}}{∣ α ∣ ^{2}} = 4 r^{2}

Multiplying by

∣ α ∣^{2}

to clear the denominators, we obtain:

1 + ∣ z_{0} ∣^{2} ∣ α ∣^{2} - ∣ α ∣^{2} - ∣ z_{0} ∣^{2} + r^{2} = 4 r^{2} ∣ α ∣^{2}

Grouping the terms leads to:

(∣ α ∣^{2} - 1) (∣ z_{0} ∣^{2} - 1) + r^{2} = 4 r^{2} ∣ α ∣^{2}

Given the condition

2∣ z_{0} ∣^{2} = r^{2} + 2

, we have

∣ z_{0} ∣^{2} - 1 = \frac{r ^{2}}{2}

. Substituting this into our equation:

(∣ α ∣^{2} - 1) (\frac{r ^{2}}{2}) + r^{2} = 4 r^{2} ∣ α ∣^{2}

Dividing by

r^{2}

(since $r eq 0$) and multiplying by 2, we get:

∣ α ∣^{2} - 1 + 2 = 8∣ α ∣^{2}

This simplifies to

1 = 7∣ α ∣^{2}

. Therefore, the final result is:

$∣ α ∣ = \frac{1}{7}$

Common Traps & FAQs

Wait, why can we assume

α

and

1/ \overset{α}{ˉ}

are collinear with the origin?

Great observation! Remember that

\overset{α}{ˉ} = ∣ α ∣^{2} / α

. Therefore,

1/ \overset{α}{ˉ} = α /∣ α ∣^{2}

. Since

∣ α ∣^{2}

is a positive real number,

1/ \overset{α}{ˉ}

is just a positive scalar multiple of

α

. In the complex plane, multiplying a vector by a positive scalar doesn't change its direction, only its magnitude. Thus, they must lie on the same ray starting from the origin.

Why do we use

∣ z ∣^{2} = z \overset{z}{ˉ}

instead of just working with coordinates

x

and

y

You *could* use coordinates, but it would be a nightmare of algebra! Using

∣ z ∣^{2} = z \overset{z}{ˉ}

is the 'pro' way to handle complex geometry. It keeps the expressions symmetric and allows us to use the properties of complex numbers directly, which often leads to beautiful cancellations that coordinate geometry would hide.

Is it possible for

r = 0

in this problem?

Excellent question! If

r = 0

, the circles would collapse into a single point,

z_{0}

. However, the problem defines two distinct circles with radii

r

and

2 r

. If

r = 0

, the radius of the outer circle would also be zero, which contradicts the geometric setup. Therefore, we can safely assume

r \neq = 0

and divide by

r^{2}

during our simplification.

Select Answer:

Visualized Solution

Visualizing the Concentric Circles

The Collinearity of $α$ and $\frac{1}{α ˉ}$

Setting up the Circle Equations

Expanding $∣ α - z_{0} ∣^{2} = r^{2}$

Expanding $∣ \frac{1}{α ˉ} - z_{0} ∣^{2} = 4 r^{2}$

Introducing the Substitution Variable $K$

Eliminating the Variable $K$

Simplifying the Combined Equation

Applying the Condition $2∣ z_{0} ∣^{2} = r^{2} + 2$

Solving the Algebraic Equation

Conclusion and Key Takeaways

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Algebraic Dance

The Final Reveal

Similar Questions

For $α, β, z \in C$ and $λ > 1$ , if $λ - 1$ is the radius of the circle $∣ z - α ∣^{2} + ∣ z - β ∣^{2} = 2 λ$ , then $∣ α - β ∣$ is equal to

Let the complex numbers $α$ and $\frac{1}{α ˉ}$ lie on the circles $∣ z - z_{0} ∣^{2} = 4$ and $∣ z - z_{0} ∣^{2} = 16$ respectively, where $z_{0} = 1 + i$ . Then, the value of $100∣ α ∣^{2}$ is.

Let $z$ be a complex number such that $∣ \frac{z - 2 i}{z + i} ∣ = 2, z \neq = - i$ . Then $z$ lies on the circle of radius 2 and centre

If $\frac{z - α}{z + α}$ ( $α \in R$ ) is a purely imaginary number and $∣ z ∣ = 2$ , then a value of $α$ is :

If $z_{1}, z_{2}$ are two distinct complex number such that $\frac{z _{1} - 2 z _{2}}{\frac{1}{2} - z _{1} z ˉ _{2}} = 2$ , then

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

Let $z$ be a complex number such that $∣ z ∣ = 1$ . If $\frac{2 + k ^{2} z}{k + z ˉ} = k z, k \in R$ , then the maximum distance of $k + i k^{2}$ from the circle $∣ z - (1 + 2 i) ∣ = 1$ is:

If the locus of $z \in C$ , such that $Re (\frac{z - 1}{2 z + i}) + Re (\frac{z ˉ - 1}{2 z ˉ - i}) = 2$ is a circle of radius $r$ and center $(a, b)$ then $\frac{15 ab}{r ^{2}}$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Concentric Circles

The Collinearity of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } αˉ1​

Setting up the Circle Equations

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣α−z0​∣2=r2

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣αˉ1​−z0​∣2=4r2

Introducing the Substitution Variable .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K

Eliminating the Variable .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } K

Simplifying the Combined Equation

Applying the Condition .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2∣z0​∣2=r2+2

Solving the Algebraic Equation

Conclusion and Key Takeaways

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Algebraic Dance

The Final Reveal

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For α,β,z∈C and λ>1, if λ−1​ is the radius of the circle ∣z−α∣2+∣z−β∣2=2λ, then ∣α−β∣ is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the complex numbers α and αˉ1​ lie on the circles ∣z−z0​∣2=4 and ∣z−z0​∣2=16 respectively, where z0​=1+i. Then, the value of 100∣α∣2 is.

.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+iz−2i​∣=2,z=−i. Then z lies on the circle of radius 2 and centre

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z+αz−α​ (α∈R) is a purely imaginary number and ∣z∣=2, then a value of α is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If z1​,z2​ are two distinct complex number such that ​21​−z1​zˉ2​z1​−2z2​​​=2, then

.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; } Let z be a complex number such that ∣z∣=1. If k+zˉ2+k2z​=kz,k∈R, then the maximum distance of k+ik2 from the circle ∣z−(1+2i)∣=1 is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the locus of z∈C, such that Re(2z+iz−1​)+Re(2zˉ−izˉ−1​)=2 is a circle of radius r and center (a,b) then r215ab​ is equal to :

The Collinearity of $α$ and $\frac{1}{α ˉ}$

Expanding $∣ α - z_{0} ∣^{2} = r^{2}$

Expanding $∣ \frac{1}{α ˉ} - z_{0} ∣^{2} = 4 r^{2}$

Introducing the Substitution Variable $K$

Eliminating the Variable $K$

Applying the Condition $2∣ z_{0} ∣^{2} = r^{2} + 2$

For $α, β, z \in C$ and $λ > 1$ , if $λ - 1$ is the radius of the circle $∣ z - α ∣^{2} + ∣ z - β ∣^{2} = 2 λ$ , then $∣ α - β ∣$ is equal to

Let the complex numbers $α$ and $\frac{1}{α ˉ}$ lie on the circles $∣ z - z_{0} ∣^{2} = 4$ and $∣ z - z_{0} ∣^{2} = 16$ respectively, where $z_{0} = 1 + i$ . Then, the value of $100∣ α ∣^{2}$ is.

Let $z$ be a complex number such that $∣ \frac{z - 2 i}{z + i} ∣ = 2, z \neq = - i$ . Then $z$ lies on the circle of radius 2 and centre

If $\frac{z - α}{z + α}$ ( $α \in R$ ) is a purely imaginary number and $∣ z ∣ = 2$ , then a value of $α$ is :

If $z_{1}, z_{2}$ are two distinct complex number such that $\frac{z _{1} - 2 z _{2}}{\frac{1}{2} - z _{1} z ˉ _{2}} = 2$ , then

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

Let $z$ be a complex number such that $∣ z ∣ = 1$ . If $\frac{2 + k ^{2} z}{k + z ˉ} = k z, k \in R$ , then the maximum distance of $k + i k^{2}$ from the circle $∣ z - (1 + 2 i) ∣ = 1$ is:

If the locus of $z \in C$ , such that $Re (\frac{z - 1}{2 z + i}) + Re (\frac{z ˉ - 1}{2 z ˉ - i}) = 2$ is a circle of radius $r$ and center $(a, b)$ then $\frac{15 ab}{r ^{2}}$ is equal to :