Animated Solution for Mathematics - Complex Numbers: The area of the triangle with vertices $A(z),B(iz)$ and $C(z+iz)$ is :

Imagine you are standing at the origin of the Argand plane. You have a complex number $z$, which we can visualize as a vector $\overrightarrow{OA}$ pointing from the origin to a point $A$.

In the world of complex numbers, algebra is not just about symbols; it is about geometry. When we multiply $z$ by $i$, we are performing a physical transformation. Multiplying by $i$ rotates the vector $\overrightarrow{OA}$ by exactly $90^{\circ }$ counter-clockwise.

This gives us point $B$, representing $iz$. Because the magnitude of $i$ is $1$, the length of $\overrightarrow{OB}$ is identical to the length of $\overrightarrow{OA}$, which is $|z|$. We have now established two sides of a quadrilateral: $\overrightarrow{OA}$ and $\overrightarrow{OB}$, both of length $|z|$, meeting at a perfect right angle.

Now, let us introduce the third vertex, $C(z+iz)$. In the language of vectors, this is simply the sum of $\overrightarrow{OA}$ and $\overrightarrow{OB}$.

By the parallelogram law of vector addition, the point $C$ is the fourth vertex of the parallelogram $OACB$. But wait—look closer. We have a parallelogram where adjacent sides are equal ($|z|$) and the angle between them is $90^{\circ }$.

A parallelogram with equal adjacent sides and a right angle is, by definition, a square. This realization is the key to the entire problem. We are not dealing with an arbitrary triangle; we are dealing with a triangle embedded within a square.

The question asks for the area of $△ABC$. If you sketch this out, you will see that $△ABC$ is a right-angled triangle.

Because $OACB$ is a square, the angle at $C$ is $90^{\circ }$. To find the area, we need the lengths of the base and height, which are the sides $AC$ and $BC$.

The length of $AC$ is the distance between $C(z+iz)$ and $A(z)$. Mathematically, this is:

Using the property that the modulus of a product is the product of the moduli, we get:

Similarly, the length of $BC$ is the distance between $C(z+iz)$ and $B(iz)$, which is:

We have found that both $AC$ and $BC$ are equal to $|z|$. Since $△ABC$ is a right-angled triangle at $C$, its area is simply $\frac{1}{2}\times \text{base}\times \text{height}$.

Substituting our values, we get:

It is a beautiful, clean result. We started with a seemingly abstract complex number problem and ended with a simple geometric area. This is the power of visualization in JEE Advanced mathematics; never fear the complex, embrace the geometry behind it.