Complex Numbers – Geometry of Triangles – JEE Main 28 Jan 2025 Shift 1

Solution:

Given $z_1 = \sqrt{3} + 2\sqrt{2}i$.
The relation between magnitudes is $\frac{|z_2|}{|z_1|} = \frac{1}{\sqrt{3}}$.
Given $\arg\left(\frac{z_2}{z_1}\right) = \frac{\pi}{6}$.
Using the rotation formula:
$$ z_2 = \frac{|z_2|}{|z_1|} \cdot z_1 \cdot e^{i(\pi/6)} $$
$$ z_2 = \frac{1}{\sqrt{3}} (\sqrt{3} + 2\sqrt{2}i) \left( \frac{\sqrt{3}}{2} + \frac{i}{2} \right) $$
$$ z_2 = \frac{1}{\sqrt{3}} \left( \frac{3}{2} + \frac{\sqrt{3}i}{2} + \sqrt{6}i – \sqrt{2} \right) $$
(Note: The PDF solution calculates $(z_1 – z_2)$ to check the side lengths).

Let’s find side $AB = |z_1 – z_2|$.
Using the Cosine Rule in $\Delta ABO$:
$$ AB^2 = OA^2 + OB^2 – 2(OA)(OB)\cos(60^\circ \text{ or } 30^\circ?) $$
Wait, the angle of rotation is $\pi/6 = 30^\circ$.
$$ AB^2 = |z_1|^2 + |z_2|^2 – 2|z_1||z_2|\cos(30^\circ) $$
Substitute $|z_2| = \frac{|z_1|}{\sqrt{3}}$:
$$ AB^2 = |z_1|^2 + \frac{|z_1|^2}{3} – 2|z_1|\frac{|z_1|}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} $$
$$ AB^2 = |z_1|^2 \left( 1 + \frac{1}{3} – 1 \right) = \frac{|z_1|^2}{3} $$
$$ AB = \frac{|z_1|}{\sqrt{3}} $$

Now compare the three sides:
1. $OA = |z_1|$
2. $OB = |z_2| = \frac{|z_1|}{\sqrt{3}}$
3. $AB = \frac{|z_1|}{\sqrt{3}}$

Since $OB = AB$, the triangle is **Isosceles**.

Now check for the type of angle.
The sides are proportional to $\sqrt{3} : 1 : 1$.
Using the Cosine rule to find angle at B (opposite to largest side OA):
$$ \cos B = \frac{1^2 + 1^2 – (\sqrt{3})^2}{2(1)(1)} = \frac{2 – 3}{2} = -\frac{1}{2} $$
$$ B = 120^\circ = \frac{2\pi}{3} $$
Since one angle is $120^\circ$, it is an **Obtuse Angled Isosceles Triangle**.

Ans. (4)