JEE Main 2023 (29 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Complex Numbers: For two non-zero complex number $z_{1}$ and $z_{2}$ , if $Re (z_{1} z_{2}) = 0$ and $Re (z_{1} + z_{2}) = 0$ , then which of the following are possible? \\ (A) $Im (z_{1}) > 0$ and $Im (z_{2}) > 0$ \\ (B) $Im (z_{1}) < 0$ and $Im (z_{2}) > 0$ \\ (C) $Im (z_{1}) > 0$ and $Im (z_{2}) < 0$ \\ (D) $Im (z_{1}) < 0$ and $Im (z_{2}) < 0$ \\ Choose the correct answer from the options given below:

Q: Wait, why can't $x_1$ be zero? If $x_1$ is zero, doesn't the equation still hold?

That is a fantastic question! Let's look at the logic. If $x_1 = 0$, then our condition $\text{Re}(z_1 + z_2) = 0$ forces $x_2 = -x_1 = 0$. Now, look at our product equation: $y_1 y_2 = -x_1^2$. If $x_1 = 0$, then $y_1 y_2 = 0$. This implies that either $y_1 = 0$ or $y_2 = 0$. If $y_1 = 0$, then $z_1 = x_1 + iy_1 = 0 + i(0) = 0$. But the problem explicitly states that $z_1$ and $z_2$ are non-zero complex numbers! Therefore, $x_1$ cannot be zero, because it would force one of our complex numbers to collapse into the origin.

Q: Why does $y_1 y_2 < 0$ necessarily mean they have opposite signs?

Think about the properties of multiplication for real numbers. If the product of two real numbers is negative, it is mathematically impossible for them to have the same sign. If both were positive, their product would be positive. If both were negative, their product would also be positive (a negative times a negative is a positive). The only way to get a negative result is to have one positive and one negative factor. Thus, $y_1$ and $y_2$ must have opposite signs.

Q: Can we solve this using polar form $z = r e^{i\theta}$ instead of $x + iy$?

You absolutely can, but it is like taking the scenic route through a mountain range when there is a tunnel right in front of you! Using polar form would lead you to trigonometric equations involving $\cos(\theta_1 + \theta_2) = 0$. While that is a valid and elegant approach, it requires more steps and is prone to calculation errors. In JEE Advanced, the best strategy is often the one that minimizes algebraic complexity. Stick to the Cartesian form $x + iy$ for this specific problem; it is the most direct path to the solution.

Select Answer:

Visualized Solution

Defining $z_{1}$ and $z_{2}$

Let $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$
Given $z_{1}, z_{2} \neq = 0$ , so $(x_{1}, y_{1}) \neq = (0, 0)$ and $(x_{2}, y_{2}) \neq = (0, 0)$

Analyzing $Re (z_{1} + z_{2}) = 0$

Condition 1: $Re (z_{1} + z_{2}) = 0$
$\Rightarrow x_{1} + x_{2} = 0$
$\Rightarrow x_{2} = - x_{1}$

Analyzing $Re (z_{1} z_{2}) = 0$

Condition 2: $Re (z_{1} z_{2}) = 0$
Product $z_{1} z_{2} = (x_{1} + i y_{1}) (x_{2} + i y_{2})$

Expanding the Product

$= (x_{1} x_{2} - y_{1} y_{2}) + i (x_{1} y_{2} + x_{2} y_{1})$
$\Rightarrow Re (z_{1} z_{2}) = x_{1} x_{2} - y_{1} y_{2} = 0$

Substituting $x_{2} = - x_{1}$

Substitute $x_{2} = - x_{1}$ into $x_{1} x_{2} - y_{1} y_{2} = 0$ :
$x_{1} (- x_{1}) - y_{1} y_{2} = 0$

The Core Relationship

$- x_{1}^{2} - y_{1} y_{2} = 0$
$\Rightarrow y_{1} y_{2} = - x_{1}^{2}$

Checking the Non-Zero Constraint

If $x_{1} = 0$ , then $y_{1} y_{2} = 0$ .
This implies $y_{1} = 0$ or $y_{2} = 0$ , making $z_{1} = 0$ or $z_{2} = 0$ .
This contradicts $z_{1}, z_{2} \neq = 0$ , so $x_{1} \neq = 0$ .

Determining the Signs

Since $x_{1} \neq = 0$ , we have $x_{1}^{2} > 0$ .
Therefore, $y_{1} y_{2} = - x_{1}^{2} < 0$ .
This implies $y_{1}$ and $y_{2}$ must have opposite signs.

Evaluating the Options

Check possibilities:
(B) $Im (z_{1}) < 0$ and $Im (z_{2}) > 0$ (Opposite signs - Possible)
(C) $Im (z_{1}) > 0$ and $Im (z_{2}) < 0$ (Opposite signs - Possible)

00:00 / 00:00

The Sigma Insight: Algebraic Operations on Complex Numbers

Solution Diagram

Analyzing the Setup

We represent the two non-zero complex numbers as

z_{1} = x_{1} + i y_{1}

and

z_{2} = x_{2} + i y_{2}

. Since $z_1, z_2 eq 0$, we know that $(x_1, y_1) eq (0,0)$ and $(x_2, y_2) eq (0,0)$.

The first condition given is

Re (z_{1} + z_{2}) = 0

. Expanding the sum, we have:

z_{1} + z_{2} = (x_{1} + x_{2}) + i (y_{1} + y_{2})

The real part is

x_{1} + x_{2} = 0

, which implies:

x_{2} = - x_{1}

The Master Equation

The second condition is

Re (z_{1} z_{2}) = 0

. Expanding the product of the two complex numbers:

z_{1} z_{2} = (x_{1} + i y_{1}) (x_{2} + i y_{2}) = (x_{1} x_{2} - y_{1} y_{2}) + i (x_{1} y_{2} + x_{2} y_{1})

Setting the real part to zero, we obtain:

x_{1} x_{2} - y_{1} y_{2} = 0

Substituting

x_{2} = - x_{1}

into this equation yields:

x_{1} (- x_{1}) - y_{1} y_{2} = 0

- x_{1}^{2} - y_{1} y_{2} = 0

y_{1} y_{2} = - x_{1}^{2}

Final Calculation and Interpretation

We must determine the sign of the product

y_{1} y_{2}

. If

x_{1} = 0

, then

x_{2} = 0

, which forces

y_{1}

y_{2}

to be zero for the product to be non-zero, contradicting the premise that $z_1, z_2 eq 0$.

Thus, $x_1 eq 0$, which implies

x_{1}^{2} > 0

. Consequently,

- x_{1}^{2} < 0

This leads to the definitive conclusion:

y_{1} y_{2} < 0

The imaginary parts of $z_{1}$ and $z_{2}$ must have opposite signs. This geometric relationship confirms that the complex numbers are positioned such that their imaginary components are strictly of opposite polarity.

Common Traps & FAQs

Wait, why can't

x_{1}

be zero? If

x_{1}

is zero, doesn't the equation still hold?

That is a fantastic question! Let's look at the logic. If

x_{1} = 0

, then our condition

Re (z_{1} + z_{2}) = 0

forces

x_{2} = - x_{1} = 0

. Now, look at our product equation:

y_{1} y_{2} = - x_{1}^{2}

. If

x_{1} = 0

, then

y_{1} y_{2} = 0

. This implies that either

y_{1} = 0

y_{2} = 0

. If

y_{1} = 0

, then

z_{1} = x_{1} + i y_{1} = 0 + i (0) = 0

. But the problem explicitly states that

z_{1}

and

z_{2}

are non-zero complex numbers! Therefore,

x_{1}

cannot be zero, because it would force one of our complex numbers to collapse into the origin.

Why does

y_{1} y_{2} < 0

necessarily mean they have opposite signs?

Think about the properties of multiplication for real numbers. If the product of two real numbers is negative, it is mathematically impossible for them to have the same sign. If both were positive, their product would be positive. If both were negative, their product would also be positive (a negative times a negative is a positive). The only way to get a negative result is to have one positive and one negative factor. Thus,

y_{1}

and

y_{2}

must have opposite signs.

Can we solve this using polar form

z = r e^{i θ}

instead of

x + i y

You absolutely can, but it is like taking the scenic route through a mountain range when there is a tunnel right in front of you! Using polar form would lead you to trigonometric equations involving

cos (θ_{1} + θ_{2}) = 0

. While that is a valid and elegant approach, it requires more steps and is prone to calculation errors. In JEE Advanced, the best strategy is often the one that minimizes algebraic complexity. Stick to the Cartesian form

x + i y

for this specific problem; it is the most direct path to the solution.

Select Answer:

Visualized Solution

Defining $z_{1}$ and $z_{2}$

Analyzing $Re (z_{1} + z_{2}) = 0$

Analyzing $Re (z_{1} z_{2}) = 0$

Expanding the Product

Substituting $x_{2} = - x_{1}$

The Core Relationship

Checking the Non-Zero Constraint

Determining the Signs

Evaluating the Options

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Master Equation

Final Calculation and Interpretation

Similar Questions

Let $z_{1}$ and $z_{2}$ be any two non-zero complex numbers such that $3∣ z_{1} ∣ = 4∣ z_{2} ∣$ . If $z = \frac{3 z _{1}}{2 z _{2}} + \frac{2 z _{2}}{3 z _{1}}$ then :

$z \neq = 0$ is a complex number

List-I

List-II

Let $a, b, x$ and $y$ be real numbers such that $a - b = 1$ and $y \neq = 0$ . If the complex number $z = x + i y$ satisfies $Im (\frac{a z + b}{z + 1}) = y$ , then which of the following is(are) possible value(s) of $x$ ?

Let $S = {z \in C : z^{2} + \overset{z}{ˉ} = 0}$ . Then $\sum_{z \in S} (Re (z) + Im (z))$ is equal to

Let $S = {z = x + i y : \frac{2 z - 3 i}{4 z + 2 i} is a real number}$ . Then which of the following is NOT correct?

Let $S$ be the set of all complex numbers $z$ satisfying $∣ z^{2} + z + 1∣ = 1$ . Then which of the following statements is/are TRUE?

Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a = z^{2} + z + 1$ is real. Then $a$ cannot take the value

Select Answer:

Visualized Solution

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z2​

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Re(z1​+z2​)=0

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Re(z1​z2​)=0

Expanding the Product

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2​=−x1​

The Core Relationship

Checking the Non-Zero Constraint

Determining the Signs

Evaluating the Options

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

The Master Equation

Final Calculation and Interpretation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let z1​ and z2​ be any two non-zero complex numbers such that 3∣z1​∣=4∣z2​∣. If z=2z2​3z1​​+3z1​2z2​​ then :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z=0 is a complex number

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={z∈C:z2+zˉ=0}. Then ∑z∈S​(Re(z)+Im(z)) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S={z=x+iy:4z+2i2z−3i​ is a real number}. Then which of the following is NOT correct?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S be the set of all complex numbers z satisfying ∣z2+z+1∣=1. Then which of the following statements is/are TRUE?

.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 the imaginary part of z is non-zero and a=z2+z+1 is real. Then a cannot take the value

Defining $z_{1}$ and $z_{2}$

Analyzing $Re (z_{1} + z_{2}) = 0$

Analyzing $Re (z_{1} z_{2}) = 0$

Substituting $x_{2} = - x_{1}$

Let $z_{1}$ and $z_{2}$ be any two non-zero complex numbers such that $3∣ z_{1} ∣ = 4∣ z_{2} ∣$ . If $z = \frac{3 z _{1}}{2 z _{2}} + \frac{2 z _{2}}{3 z _{1}}$ then :

$z \neq = 0$ is a complex number

Let $S = {z \in C : z^{2} + \overset{z}{ˉ} = 0}$ . Then $\sum_{z \in S} (Re (z) + Im (z))$ is equal to

Let $S = {z = x + i y : \frac{2 z - 3 i}{4 z + 2 i} is a real number}$ . Then which of the following is NOT correct?

Let $S$ be the set of all complex numbers $z$ satisfying $∣ z^{2} + z + 1∣ = 1$ . Then which of the following statements is/are TRUE?

Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a = z^{2} + z + 1$ is real. Then $a$ cannot take the value