Animated Solution for Mathematics - Complex Numbers: The area of the polygon, whose vertices are the non-real roots of the equation $\bar{z}=i{z}^2$ is :

My dear student, welcome to a journey through the elegant world of complex numbers. Today, we are not just solving an equation; we are uncovering a geometric secret hidden within the algebra.

We are given the equation $\bar{z}=i{z}^2$. At first glance, this looks like a standard algebraic problem, but I want you to see it as a map.

The presence of $\bar{z}$ alongside $z^2$ is a signal—a call to use the most powerful tool in our complex number toolkit: the modulus.

When we are lost in the woods of complex variables, the modulus is our compass. Let us take the modulus of both sides:

Now, remember your properties! The modulus of a conjugate is the same as the modulus of the number itself, so $|\bar{z}|=|z|$.

On the right side, the modulus of a product is the product of the moduli, so $|i{z}^2|=|i|\cdot |z{|}^2$. Since $|i|=1$, our equation simplifies beautifully to:

This is our 'Aha!' moment. Rearranging this, we get $|z|(|z|-1)=0$.

This tells us that either $|z|=0$ or $|z|=1$. We discard $z=0$ because it is a real root, and we are hunting for the non-real vertices of a polygon.

Thus, we are left with the profound realization that all our roots must lie on the unit circle, where $|z|=1$.

Now that we know our roots live on the unit circle, we can use the property $z\bar{z}=|z{|}^2=1$, which means $\bar{z}=\frac{1}{z}$.

Let us substitute this back into our original equation:

Multiplying both sides by $z$, we arrive at the clean, powerful equation $1=i{z}^3$, or:

Since $\frac{1}{i}=-i$, we are solving $z^3=-i$. This is the heart of the problem.

We are looking for the three cube roots of $-i$. In the complex plane, the roots of $z^n=c$ are always vertices of a regular $n$-gon. Here, $n=3$, so our roots must form an equilateral triangle.

Let us find these roots. We know $-i$ can be written as $e^{i(3\pi /2)}$. The roots are given by:

This gives us the following values:

Plotting these, we see the points $(0,1)$, $(-\frac{\sqrt{3}}{2},-\frac{1}{2})$, and $(\frac{\sqrt{3}}{2},-\frac{1}{2})$. These are the vertices of our triangle.

The side length $a$ is the distance between $z_2$ and $z_3$, which is simply the difference in their real parts:

We have reached the summit. The area of an equilateral triangle with side length $a$ is given by the formula:

Substituting our side length $a=\sqrt{3}$, we get:

There it is! The beauty of this problem lies in how the algebra of complex numbers seamlessly transitions into the geometry of the plane.

You started with a confusing equation and ended with a perfect equilateral triangle. The final area is $\frac{3\sqrt{3}}{4}$.