Let O be the origin, the point A be z1 = √3 + 2√2i, the point B(z2) be such that √3|z2| = |z1| and a

Q: Wait, why can't we just use the distance formula between $z_{1}$ and $z_{2}$? Why do we need the Cosine Rule?

That is a great question! While you could try to find the Cartesian coordinates of $z_{2}$ using polar form $z_{2} = |z_{2}|e^{i\text{arg}(z_{2})}$, it is much more tedious. The Cosine Rule is a powerful shortcut because the problem already gives you the lengths of two sides and the angle between them. In JEE Advanced, recognizing these geometric shortcuts saves precious time!

Q: How do I know if the triangle is obtuse or acute? Is there a quick check?

Absolutely! For any triangle with sides $a, b, c$ where $c$ is the longest side, calculate $c^{2}$ and compare it to $a^{2} + b^{2}$. If $c^{2} > a^{2} + b^{2}$, the angle opposite to $c$ is obtuse. If $c^{2} 7.33$, it is obtuse.

Q: Why is the angle between $OA$ and $OB$ exactly $\frac{\pi}{6}$?

Remember that the argument of a complex number, $\text{arg}(z)$, represents the angle the vector makes with the positive real axis. If $\text{arg}(z_{2}) = \text{arg}(z_{1}) + \frac{\pi}{6}$, then the difference $\text{arg}(z_{2}) - \text{arg}(z_{1}) = \frac{\pi}{6}$ is the angle between the two vectors $OA$ and $OB$. It is a direct geometric interpretation of the difference in their arguments.

Question

Let O be the origin, the point A be $z_{1}=\sqrt{3}+2\sqrt{2}i,$ the point $B(z_{2})$ be such that $\sqrt{3}|z_{2}|=|z_{1}|$ and $arg(z_{2})=arg(z_{1})+\frac{\pi}{6}$. Then

Options:

(A) area of triangle ABO is $\frac{11}{\sqrt{3}}$

(B) ABO is an obtuse angled isosceles triangle

(C) area of triangle ABO is $\frac{11}{4}$

(D) ABO is a scalene triangle

Sigma Percentile · Accepted Answer

### The Geometry of Complex Numbers: A Journey Through $	riangle ABO$

Imagine you are standing at the origin of the complex plane. You have two points, $A$ and $B$, floating in this two-dimensional space.

At first glance, this problem might look like a dry algebraic exercise, but it is actually a beautiful geometric puzzle waiting to be solved. Let us break it down together.

### Phase 1: Locating Point A

We begin with point $A$, defined by the complex number $z_{1} = \sqrt{3} + 2\sqrt{2}i$. To understand its position, we need its distance from the origin, which is its modulus, $|z_{1}|$.

Using the standard formula $|z| = \sqrt{	ext{Re}(z)^{2} + 	ext{Im}(z)^{2}}$, we calculate:

$$|z_{1}| = \sqrt{(\sqrt{3})^{2} + (2\sqrt{2})^{2}} = \sqrt{3 + 8} = \sqrt{11}$$

So, the length of the segment $OA$ is $\sqrt{11}$. This is our anchor.

### Phase 2: The Transformation to Point B

Now, consider point $B$. We are given two crucial pieces of information: $\sqrt{3}|z_{2}| = |z_{1}|$ and $	ext{arg}(z_{2}) = 	ext{arg}(z_{1}) + \frac{\pi}{6}$.

The first tells us the scaling:

$$|z_{2}| = \frac{|z_{1}|}{\sqrt{3}} = \sqrt{\frac{11}{3}}$$

The second tells us the rotation: the vector $OB$ is rotated by $\frac{\pi}{6}$ (or $30^{\circ}$) relative to $OA$. Geometrically, this means the angle $\angle AOB$ is exactly $30^{\circ}$.

### Phase 3: The Power of the Cosine Rule

We now have a triangle $	riangle AOB$ where we know two sides ($OA = \sqrt{11}$ and $OB = \sqrt{\frac{11}{3}}$) and the included angle ($\angle AOB = 30^{\circ}$). To find the third side, $AB$, we reach for the Law of Cosines:

$$AB^{2} = OA^{2} + OB^{2} - 2(OA)(OB)\cos(30^{\circ})$$

Substituting our values:

$$AB^{2} = 11 + \frac{11}{3} - 2(\sqrt{11})\left(\sqrt{\frac{11}{3}}ight)\left(\frac{\sqrt{3}}{2}ight)$$

Watch the magic happen as we simplify. The $2$ in the numerator and denominator cancels out. The $\sqrt{3}$ in the denominator of the middle term cancels with the $\sqrt{3}$ from the cosine.

We are left with:

$$AB^{2} = 11 + \frac{11}{3} - 11 = \frac{11}{3}$$

Thus, $AB = \sqrt{\frac{11}{3}}$.

### Phase 4: The Final Classification

Look at our side lengths: $OA = \sqrt{11}$, $OB = \sqrt{\frac{11}{3}}$, and $AB = \sqrt{\frac{11}{3}}$. Since $OB = AB$, the triangle is clearly isosceles.

But is it obtuse? We check the square of the longest side, $OA^{2} = 11$, against the sum of the squares of the other two sides:

$$OB^{2} + AB^{2} = \frac{11}{3} + \frac{11}{3} = \frac{22}{3} \approx 7.33$$

Because $11 > 7.33$, the angle opposite $OA$ must be obtuse. We have arrived at our destination: **$	riangle ABO$ is an obtuse-angled isosceles triangle.**

Animated Solution for Mathematics - Complex Numbers: Let O be the origin, the point A be $z_{1} = 3 + 22 i,$ the point $B (z_{2})$ be such that $3 ∣ z_{2} ∣ = ∣ z_{1} ∣$ and $a r g (z_{2}) = a r g (z_{1}) + \frac{π}{6}$ . Then

Select Answer:

Visualized Solution

Visualizing Point A in Complex Plane

Calculating Distance $O A$

Finding Distance $O B$

Determining Angle $∠ A O B$

The Cosine Rule

Substituting Values

Calculating Side $A B$

Identifying the Triangle Type

Checking for Obtuse Angle

Final Conclusion

The Sigma Insight: Geometrical Applications of Complex Numbers

The Geometry of Complex Numbers

Phase 1

Phase 2

Phase 3

Phase 4

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