Let z1 and z2 be complex numbers such that z1 ≠ z2 and |z1| = |z2|. If z1 has positive real part and

Q: Why do we consider the case where the ratio is zero? Isn't that just a trivial case?

In JEE Advanced, the 'trivial' cases are often the traps! If $z_1 = -z_2$, the denominator $z_1 - z_2$ is $2z_1$, which is non-zero, but the numerator $z_1 + z_2$ becomes zero. Always check the boundaries of your conditions. The problem states $z_1 \neq z_2$, but it does not forbid $z_1 = -z_2$.

Q: Why is the result purely imaginary? I don't see the connection to the rhombus.

Think of the vectors $z_1$ and $z_2$ as sides of a rhombus. The sum $z_1+z_2$ is the major diagonal, and the difference $z_1-z_2$ is the minor diagonal. In a rhombus, diagonals are perpendicular. A ratio of two perpendicular vectors in the complex plane is always purely imaginary. The algebra confirms what the geometry dictates!

Q: Can I just substitute $z_1 = 1+i$ and $z_2 = 1-i$ to check?

That is a great strategy for a quick check, but be careful! $1+i$ and $1-i$ have the same modulus, but $Re(z_1) > 0$ and $Im(z_2) < 0$ are satisfied. However, you must ensure your chosen values satisfy *all* constraints. If you only test one case, you might miss the 'zero' possibility. Always back up your specific test cases with the general algebraic proof.

Question

Let $z_1$ and $z_2$ be complex numbers such that $z_1 
eq z_2$ 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

Options:

(A) zero

(B) real and positive

(C) real and negative

(D) purely imaginary

Sigma Percentile · Accepted Answer

### Analyzing the Geometric Setup

We are given two complex numbers, $z_1$ and $z_2$, such that their magnitudes are equal: $|z_1| = |z_2| = r$. Geometrically, this implies that both $z_1$ and $z_2$ lie on a circle of radius $r$ centered at the origin.

The constraints $Re(z_1) > 0$ and $Im(z_2) < 0$ restrict $z_1$ to the right half-plane and $z_2$ to the lower half-plane. We are tasked with evaluating the ratio:

$$Z = \frac{z_1 + z_2}{z_1 - z_2}$$

### The Exponential Transformation

To simplify this expression, we represent the complex numbers in polar form: $z_1 = re^{i	heta_1}$ and $z_2 = re^{i	heta_2}$. Substituting these into the ratio, the radius $r$ cancels out entirely:

$$Z = \frac{re^{i	heta_1} + re^{i	heta_2}}{re^{i	heta_1} - re^{i	heta_2}} = \frac{e^{i	heta_1} + e^{i	heta_2}}{e^{i	heta_1} - e^{i	heta_2}}$$

To proceed, we factor out the average angle $e^{i(	heta_1+	heta_2)/2}$ from both the numerator and the denominator:

$$Z = \frac{e^{i(	heta_1+	heta_2)/2} \left( e^{i(	heta_1-	heta_2)/2} + e^{-i(	heta_1-	heta_2)/2} ight)}{e^{i(	heta_1+	heta_2)/2} \left( e^{i(	heta_1-	heta_2)/2} - e^{-i(	heta_1-	heta_2)/2} ight)}$$

### Simplifying via Euler's Identities

The exponential terms outside the parentheses cancel out. Applying Euler's identities, specifically $e^{ix} + e^{-ix} = 2\cos(x)$ and $e^{ix} - e^{-ix} = 2i\sin(x)$, we obtain:

$$Z = \frac{2\cos\left(\frac{	heta_1 - 	heta_2}{2}ight)}{2i\sin\left(\frac{	heta_1 - 	heta_2}{2}ight)}$$

This simplifies to the following expression:

$$Z = -i \cot\left(\frac{	heta_1 - 	heta_2}{2}ight)$$

### Geometric Interpretation and Edge Cases

Geometrically, $z_1$ and $z_2$ represent adjacent sides of a rhombus. The sum $z_1 + z_2$ and the difference $z_1 - z_2$ represent the diagonals of this rhombus, which are known to be perpendicular. A ratio of two perpendicular vectors is **purely imaginary**.

However, we must consider the edge case where the denominator $z_1 - z_2 = 0$. This occurs if $z_1 = z_2$, which is excluded by the quadrant constraints.

Another critical case is when the numerator $z_1 + z_2 = 0$, which occurs if $z_1 = -z_2$. Given the constraints, if $z_1$ is in the first quadrant, $-z_1$ lies in the third quadrant, which satisfies $Im(z_2) < 0$. In this specific scenario, the ratio becomes **zero**.

Thus, the value of the expression is **purely imaginary or zero**.

Animated Solution for Mathematics - Complex Numbers: 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

Select Answer:

Visualized Solution

Modulus Condition $∣ z_{1} ∣ = ∣ z_{2} ∣$

Position of $z_{1}$

Position of $z_{2}$

The Target Ratio

Polar Representation

Canceling the Radius

Half-Angle Factorization

Euler's Identity

Purely Imaginary Result

Geometric Interpretation

The Rhombus Property

Sum and Difference Vectors

Perpendicular Diagonals

The Zero Trap

The Sigma Insight: Geometrical Applications of Complex Numbers

Analyzing the Geometric Setup

The Exponential Transformation

Simplifying via Euler's Identities

Geometric Interpretation and Edge Cases

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