Animated Solution for Mathematics - Complex Numbers: Let $O$ be the origin and $A$ be the point $z_1=1+2i$. If $B$ is the point $z_2$, $\text{Re}(z_2)<0$, such that $OAB$ is a right angled isosceles triangle with $OB$ as hypotenuse, then which of the following is NOT true ?

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey through the complex plane. This problem is a classic JEE Advanced gem that tests your ability to bridge the gap between pure geometry and the algebraic power of complex numbers.

Imagine the complex plane with the origin $O$ at $0+0i$ and point $A$ at $z_1=1+2i$. The problem states that $△OAB$ is a right-angled isosceles triangle with $OB$ as the hypotenuse.

In any right-angled triangle, the hypotenuse is the side opposite the right angle. Therefore, the right angle must be at vertex $A$, meaning $\angle OAB=90^{\circ }$. Furthermore, because it is isosceles, the legs adjacent to the right angle must be equal in length, so $|OA|=|AB|$.

How do we find point $B(z_2)$? We could use coordinate geometry, but that is the long road. Instead, let us use the Rotation Theorem.

We want to rotate the vector $\overrightarrow{AO}$ by $90^{\circ }$ to get the vector $\overrightarrow{AB}$. In the complex plane, rotating a vector by $90^{\circ }$ is as simple as multiplying by $i$ (for counter-clockwise) or $-i$ (for clockwise).

Mathematically, we express this as:

This equation is the heart of the problem. It tells us that the vector from $A$ to $B$ is the vector from $A$ to $O$ rotated by $90^{\circ }$. The $\pm$ sign accounts for the two possible orientations of the triangle.

Let us solve for $z_2$. We have two cases to consider:

Case 1: $z_2-z_1=-i{z}_1$. This simplifies to $z_2=z_1(1-i)$.

Substituting $z_1=1+2i$, we get:

Here, $\text{Re}(z_2)=3$, which is greater than $0$. But the problem explicitly demands $\text{Re}(z_2)<0$. Thus, we must reject this case.

Case 2: $z_2-z_1=i{z}_1$. This simplifies to $z_2=z_1(1+i)$.

Substituting $z_1=1+2i$, we get:

Here, $\text{Re}(z_2)=-1$, which is indeed less than $0$. This is our valid point $B$.

Now that we have our $z_2=-1+3i$, let us verify the properties:

For the argument, $\arg z_2=\arg (-1+3i)$. Since this point is in the second quadrant, the argument is $\pi -{\tan }^{-1}(3)$.

For the expression $z_1-2z_2$:

The magnitude $|z_2|=\sqrt{(-1{)}^2+3^2}=\sqrt{10}$. Finally, for the expression $|2z_1-z_2|$:

If an option claims the value is $5$, that statement is NOT true. You see, my friend? When you master the rotation operator, complex geometry becomes a playground rather than a battlefield.