JEE Main 2023 (10 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: Let the complex number $z = x + i y$ be such that $\frac{2 z - 3 i}{2 z + i}$ is purely imaginary. If $x + y^{2} = 0$ , then $y^{4} + y^{2} - y$ is equal to

Q: Wait, why do we multiply by the conjugate of the denominator? Can't we just set the real part of the numerator to zero?

That is a classic trap! A fraction $\frac{A}{B}$ is purely imaginary if $\text{Re}(\frac{A}{B}) = 0$. This is equivalent to $\text{Re}(A \cdot \bar{B}) = 0$. If you only set $\text{Re}(A) = 0$, you are ignoring the contribution of the denominator, which is not necessarily real. Always rationalize the denominator first.

Q: Is the locus of $z$ always a circle when the expression is purely imaginary?

Yes, in this form $\frac{z - z_1}{z - z_2}$, the locus is a circle with the diameter connecting $z_1$ and $z_2$. In your problem, the expression is slightly scaled, but the circular nature remains. It is a beautiful geometric property of complex numbers!

Q: Why did we substitute $x = -y^2$ instead of solving for $y$ directly?

Solving for $y$ would lead to a complex quartic equation, which is unnecessary and prone to calculation errors. The question asks for the value of the expression $y^4 + y^2 - y$, not the individual roots. Substitution is the elegant, direct path to the answer.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. When a complex number

w = \frac{A}{B}

is purely imaginary, its real part must be zero. This is equivalent to saying that the real part of the product

A \cdot \overset{ˉ}{B}

is zero. Visualize this as the condition that the vector representing

w

lies entirely on the imaginary axis in the Argand plane.

Hint 2

Hint 2: The Arsenal. To isolate the real part of the expression

\frac{2 z - 3 i}{2 z + i}

, multiply the numerator and denominator by the complex conjugate of the denominator, which is

2 \overset{z}{ˉ} - i

. This will transform the denominator into a real number, allowing you to focus solely on the real part of the numerator.

Hint 3

Hint 3: The Execution. After rationalizing, you will arrive at the equation

4 x^{2} + 4 y^{2} - 4 y - 3 = 0

. Now, use the second given condition

x = - y^{2}

to substitute

x

in terms of

y

. This will lead you directly to the expression

y^{4} + y^{2} - y

Visualized Solution

Define $z = x + i y$

Let $z = x + i y$
Given expression: $w = \frac{2 z - 3 i}{2 z + i}$

Group Real and Imaginary Parts

Substitute $z$ : $w = \frac{2 ( x + i y ) - 3 i}{2 ( x + i y ) + i}$
Numerator: $2 x + i (2 y - 3)$
Denominator: $2 x + i (2 y + 1)$

Condition for Purely Imaginary

The expression $w$ is purely imaginary.
Therefore, its Real Part must be zero: $Re (w) = 0$
To find $Re (w)$ , multiply by the conjugate of the denominator.

Extract the Real Part

$w = \frac{2 x + i ( 2 y - 3 )}{2 x + i ( 2 y + 1 )} \times \frac{2 x - i ( 2 y + 1 )}{2 x - i ( 2 y + 1 )}$
$Re (w) = \frac{( 2 x ) ( 2 x ) + ( 2 y - 3 ) ( 2 y + 1 )}{( 2 x ) ^{2} + ( 2 y + 1 ) ^{2}} = 0$

Expand the Equation

$4 x^{2} + (2 y - 3) (2 y + 1) = 0$
$4 x^{2} + 4 y^{2} + 2 y - 6 y - 3 = 0$
$4 x^{2} + 4 y^{2} - 4 y - 3 = 0$

Simplify the Locus Equation

Divide the entire equation by $4$ :
$x^{2} + y^{2} - y - \frac{3}{4} = 0$

Apply the Second Condition

We are given a second relation: $x + y^{2} = 0$
Rearranging gives: $x = - y^{2}$

Substitute $x = - y^{2}$

Substitute $x = - y^{2}$ into $x^{2} + y^{2} - y - \frac{3}{4} = 0$
$(- y^{2})^{2} + y^{2} - y - \frac{3}{4} = 0$

Final Calculation

$(- y^{2})^{2} = y^{4}$
$y^{4} + y^{2} - y - \frac{3}{4} = 0$
$y^{4} + y^{2} - y = \frac{3}{4}$

Conclusion

The required value is $\frac{3}{4}$
Key Takeaway: $Re (w) = 0$ for purely imaginary $w$ .

00:00 / 00:00

The Sigma Insight: Geometrical Applications of Complex Numbers

Solution Diagram

Analyzing the Setup

Imagine you are standing on the Argand plane, looking at the complex number

z = x + i y

. We are given a rational expression

w = \frac{2 z - 3 i}{2 z + i}

and told it is purely imaginary.

This is not just an algebraic constraint; it is a geometric one. A complex number is purely imaginary if and only if it lies on the vertical axis, meaning its real part is zero.

The Rationalization Journey

To find the real part of

w

, we must first substitute

z = x + i y

into the expression:

w = \frac{2 ( x + i y ) - 3 i}{2 ( x + i y ) + i} = \frac{2 x + i ( 2 y - 3 )}{2 x + i ( 2 y + 1 )}

To extract the real part, we multiply the numerator and denominator by the complex conjugate of the denominator, which is

2 x - i (2 y + 1)

When we multiply, the denominator becomes a real number, specifically

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

. We only care about the numerator's real part, which arises from the product of the real parts and the product of the imaginary parts (since

i^{2} = - 1

This leads us to the equation:

4 x^{2} + (2 y - 3) (2 y + 1) = 0

Expanding this, we get

4 x^{2} + 4 y^{2} - 4 y - 3 = 0

. Dividing by

4

, we find the locus of

z

is the circle:

x^{2} + y^{2} - y - \frac{3}{4} = 0

The Intersection of Curves

We are also given the condition

x + y^{2} = 0

, which implies

x = - y^{2}

. This represents a parabola.

We are looking for the intersection of this parabola and our circle. By substituting

x = - y^{2}

into our circle equation, we get:

(- y^{2})^{2} + y^{2} - y - \frac{3}{4} = 0

This simplifies to the following polynomial:

y^{4} + y^{2} - y - \frac{3}{4} = 0

Final Calculation

The question asks for the value of

y^{4} + y^{2} - y

. By simply moving the constant to the other side, we find the answer is $\frac{3}{4}$ .

It is a perfect example of how complex algebra and geometry dance together to reveal a simple, elegant truth.

Common Traps & FAQs

Wait, why do we multiply by the conjugate of the denominator? Can't we just set the real part of the numerator to zero?

That is a classic trap! A fraction

\frac{A}{B}

is purely imaginary if

Re (\frac{A}{B}) = 0

. This is equivalent to

Re (A \cdot \overset{ˉ}{B}) = 0

. If you only set

Re (A) = 0

, you are ignoring the contribution of the denominator, which is not necessarily real. Always rationalize the denominator first.

Is the locus of

z

always a circle when the expression is purely imaginary?

Yes, in this form

\frac{z - z _{1}}{z - z _{2}}

, the locus is a circle with the diameter connecting

z_{1}

and

z_{2}

. In your problem, the expression is slightly scaled, but the circular nature remains. It is a beautiful geometric property of complex numbers!

Why did we substitute

x = - y^{2}

instead of solving for

y

directly?

Solving for

y

would lead to a complex quartic equation, which is unnecessary and prone to calculation errors. The question asks for the value of the expression

y^{4} + y^{2} - y

, not the individual roots. Substitution is the elegant, direct path to the answer.

Animated Solution for Mathematics - Complex Numbers: Let the complex number $z = x + i y$ be such that $\frac{2 z - 3 i}{2 z + i}$ is purely imaginary. If $x + y^{2} = 0$ , then $y^{4} + y^{2} - y$ is equal to

Select Answer:

Visualized Solution

Define $z = x + i y$

Group Real and Imaginary Parts

Condition for Purely Imaginary

Extract the Real Part

Expand the Equation

Simplify the Locus Equation

Apply the Second Condition

Substitute $x = - y^{2}$

Final Calculation

Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Rationalization Journey

The Intersection of Curves

Final Calculation

Similar Questions

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $ar g (z_{1} - z_{2}) = \frac{π}{4}$ and $z_{1}, z_{2}$ satisfy the equation $∣ z - 3∣ = Re (z)$ . Then the imaginary part of $z_{1} + z_{2}$ is equal to .

If $z = x + i y$ satisfies $∣ z ∣ - 2 = 0$ and $∣ z - i ∣ - ∣ z + 5 i ∣ = 0$ , then

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 $∣ α ∣ =$

If $z$ is a complex number such that $∣ z ∣ \geq 1$ , then the minimum value of $∣ z + \frac{1}{2} (3 + 4 i) ∣$ is:

Let $z$ be a complex number such that $∣ \frac{z - i}{z + 2 i} ∣ = 1$ and $∣ z ∣ = \frac{5}{2}$ . Then the value of $∣ z + 3 i ∣$ is:

If $z$ is a complex number such that $\frac{z - i}{z - 1}$ is purely imaginary, then the minimum value of $∣ z - (3 + 3 i) ∣$ is:

Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1} \neq = z_{2}$ and $∣ z_{1} ∣ = ∣ z_{2} ∣$ . If $z_{1}$ has positive real part and $z_{2}$ has negative imaginary part, then $(z_{1} + z_{2}) / (z_{1} - z_{2})$ may be

If $z_{1}, z_{2}$ are complex numbers such that $Re (z_{1}) = ∣ z_{1} - 1∣$ , $Re (z_{2}) = ∣ z_{2} - 1∣$ , and $arg (z_{1} - z_{2}) = \frac{π}{6}$ , then $Im (z_{1} + z_{2})$ is equal to

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

Let $z$ be the complex number satisfying $∣ z - 5∣ \leq 3$ and having maximum positive principal argument. Then $34 \frac{5 z - 12}{5 i z + 16}^{2}$ is equal to :

Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the complex number z=x+iy be such that 2z+i2z−3i​ is purely imaginary. If x+y2=0, then y4+y2−y is equal to

Select Answer:

Visualized Solution

Define .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=x+iy

Group Real and Imaginary Parts

Condition for Purely Imaginary

Extract the Real Part

Expand the Equation

Simplify the Locus Equation

Apply the Second Condition

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−y2

Final Calculation

Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Setup

The Rationalization Journey

The Intersection of Curves

Final Calculation

Similar Questions

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Animated Solution for Mathematics - Complex Numbers: Let the complex number $z = x + i y$ be such that $\frac{2 z - 3 i}{2 z + i}$ is purely imaginary. If $x + y^{2} = 0$ , then $y^{4} + y^{2} - y$ is equal to

Define $z = x + i y$

Substitute $x = - y^{2}$

If $z = x + i y$ satisfies $∣ z ∣ - 2 = 0$ and $∣ z - i ∣ - ∣ z + 5 i ∣ = 0$ , then

If $z$ is a complex number such that $∣ z ∣ \geq 1$ , then the minimum value of $∣ z + \frac{1}{2} (3 + 4 i) ∣$ is:

Let $z$ be a complex number such that $∣ \frac{z - i}{z + 2 i} ∣ = 1$ and $∣ z ∣ = \frac{5}{2}$ . Then the value of $∣ z + 3 i ∣$ is:

If $z$ is a complex number such that $\frac{z - i}{z - 1}$ is purely imaginary, then the minimum value of $∣ z - (3 + 3 i) ∣$ is:

If $z_{1}, z_{2}$ are complex numbers such that $Re (z_{1}) = ∣ z_{1} - 1∣$ , $Re (z_{2}) = ∣ z_{2} - 1∣$ , and $arg (z_{1} - z_{2}) = \frac{π}{6}$ , then $Im (z_{1} + z_{2})$ is equal to

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

Let $z$ be the complex number satisfying $∣ z - 5∣ \leq 3$ and having maximum positive principal argument. Then $34 \frac{5 z - 12}{5 i z + 16}^{2}$ is equal to :