Animated Solution for Mathematics - Complex Numbers: If $i{z}^3+z^2-z+i=0$, then show that $|z|=1$.

Imagine you are standing on the Argand plane, staring at the equation $i{z}^3+z^2-z+i=0$. It looks like a standard cubic, but there is a hidden elegance waiting to be uncovered.

Our mission is to prove that all its roots lie on the unit circle, meaning $|z|=1$. This is not just an algebraic exercise; it is a geometric revelation.

When we say $|z|=1$, we are describing a circle of radius 1 centered at the origin. Let us embark on this journey to see why these roots are destined to fall on this circle.

The secret to solving many complex polynomial problems lies in recognizing patterns. Look at the coefficients: $i,1,-1,i$. There is a beautiful symmetry here.

Instead of reaching for the cubic formula, let us group the terms: $(i{z}^3+z^2)+(-z+i)=0$. By grouping, we have created two smaller, more manageable pieces.

In the first group, $i{z}^3+z^2$, we can clearly see a common factor of $z^2$. Factoring it out, we get $z^2(iz+1)$.

Now, look at the second group, $-z+i$. We want it to look like $(iz+1)$ so we can factor the entire expression.

If we factor out $i$ from the second group, we get:

Perfect! The symmetry is revealed. Our equation now stands as $z^2(iz+1)+i(iz+1)=0$.

With the common term $(iz+1)$ identified, the polynomial collapses into a simple product:

This is the turning point. We have transformed a daunting cubic into two simple, solvable cases. This is the power of observation in JEE mathematics.

The First Root

The first case is $iz+1=0$. This implies $iz=-1$.

Dividing by $i$, we get $z=\frac{-1}{i}$. Multiplying the numerator and denominator by $i$, we find:

The modulus of this root is $|z|=|i|=1$. It sits perfectly on the unit circle.

The Modulus Shortcut

The second case is $z^2+i=0$, or $z^2=-i$. We could solve for $z$ using De Moivre's Theorem, but why do the extra work? We only need the modulus.

Taking the modulus of both sides, we get $|z^2|=|-i|$. Using the property $|z^n|=|z{|}^n$, this becomes $|z{|}^2=|-i|$.

Since the modulus of $-i$ is simply the distance from the origin, which is 1, we have $|z{|}^2=1$. Because the modulus must be a non-negative real number, we conclude $|z|=1$.

We have successfully shown that for all roots, $|z|=1$. Whether we found the root explicitly or used the properties of the modulus, the result is the same: the roots are bound to the unit circle.

This problem teaches us that in the world of complex numbers, symmetry and properties are often more powerful than brute-force calculation. Keep this perspective, and you will find that even the most intimidating equations have a simple, beautiful heart.