Let S be the set of all complex numbers z satisfying |z² + z + 1| = 1. Then which of the following s

Q: Why can't we just solve for $z$ directly like we do in a normal quadratic equation?

That is a fantastic question! In a standard quadratic equation, we are looking for specific roots. Here, we are looking for a *set* of points $S$ that satisfy a condition. The equation $|z^2 + z + 1| = 1$ defines a locus—a continuous path or curve in the complex plane. We aren't solving for a point; we are solving for the *boundaries* of this shape. We use inequalities because we are dealing with a region, not a single value.

Q: When do I use the Triangle Inequality versus the Reverse Triangle Inequality?

Think of it as a strategy of 'pushing' versus 'pulling'. Use the standard Triangle Inequality ($|a+b| \le |a| + |b|$) when you want to find a *lower bound* for a modulus, as it helps you 'push' the terms apart to see what the minimum value must be. Use the Reverse Triangle Inequality ($|a+b| \ge ||a| - |b||$) when you need to find an *upper bound* for a modulus, as it helps you 'pull' the terms together to constrain the maximum possible value. It is all about what you need to isolate!

Q: Why is the set $S$ infinite? I thought complex equations usually have a finite number of roots.

You are absolutely right that a quadratic equation has two roots. However, look closely at the equation: $|z^2 + z + 1| = 1$. This means $z^2 + z + 1$ can be *any* complex number on the unit circle. We can represent this as $z^2 + z + 1 = e^{i\theta}$, where $\theta$ is any real number. Since $\theta$ can take infinitely many values, for every single $\theta$, the quadratic $z^2 + z + (1 - e^{i\theta}) = 0$ yields two roots. Because there are infinitely many $\theta$ values, there are infinitely many $z$ values. The set $S$ is a continuous curve, not a finite collection of points.

Question

Let $S$ be the set of all complex numbers $z$ satisfying $|z^2 + z + 1| = 1$. Then which of the following statements is/are TRUE?

Options:

(A) $|z + \frac{1}{2}| \le \frac{1}{2}$ for all $z \in S$

(B) $|z| \le 2$ for all $z \in S$

(C) $|z + \frac{1}{2}| \ge \frac{1}{2}$ for all $z \in S$

(D) The set $S$ has exactly four elements

Sigma Percentile · Accepted Answer

### The Geometry of Complex Numbers: Unlocking the Locus

Welcome, future engineers! Today, we are not just solving an equation; we are embarking on a journey through the Argand plane. We are given the set $S$ of all complex numbers $z$ satisfying the condition $|z^2 + z + 1| = 1$.

At first glance, this looks like a standard algebra problem, but I want you to pause. Look at the equation. It is not asking for a root; it is asking for a **locus**. It is asking us to describe the shape, the boundaries, and the behavior of all complex numbers that live on this specific path.

Let us break this down, step by step, with the precision of a mathematician and the intuition of a physicist.

### Phase 1: The Art of Completing the Square

When we see a quadratic expression like $z^2 + z + 1$ inside a modulus, our first instinct should be to simplify. We want to see the 'center' of this expression.

The most elegant way to do this is by completing the square. We take our expression $z^2 + z + 1$ and rewrite it. We know that $z^2 + z + \frac{1}{4}$ is a perfect square, specifically $(z + \frac{1}{2})^2$. So, we split the constant $1$ into $\frac{1}{4} + \frac{3}{4}$.

Our equation becomes:

$$| (z + \frac{1}{2})^2 + \frac{3}{4} | = 1$$

Why did we do this? Because now, the term $(z + \frac{1}{2})$ is isolated. This is the key to unlocking the first bound. We are now looking at the modulus of the sum of two complex numbers: $(z + \frac{1}{2})^2$ and $\frac{3}{4}$.

### Phase 2: The Triangle Inequality – Our First Weapon

Now, we invoke the Triangle Inequality. This is one of the most fundamental tools in your JEE arsenal. It states that for any two complex numbers $a$ and $b$, $|a + b| \le |a| + |b|$.

Geometrically, this tells us that the length of one side of a triangle is always less than or equal to the sum of the other two sides. Applying this to our equation:

$$1 = | (z + \frac{1}{2})^2 + \frac{3}{4} | \le | (z + \frac{1}{2})^2 | + | \frac{3}{4} |$$

Since $| \frac{3}{4} | = \frac{3}{4}$, we have:

$$1 \le | z + \frac{1}{2} |^2 + \frac{3}{4}$$

Rearranging this, we get $| z + \frac{1}{2} |^2 \ge 1 - \frac{3}{4} = \frac{1}{4}$. Taking the square root, we arrive at $| z + \frac{1}{2} | \ge \frac{1}{2}$.

This is a profound result! It tells us that all complex numbers $z$ in our set $S$ must lie on or outside a circle centered at $-\frac{1}{2}$ with a radius of $\frac{1}{2}$. **Option C is confirmed as TRUE.**

### Phase 3: The Reverse Triangle Inequality – The Secret Weapon

Now, let us tackle the bound for $|z|$. We need to find the upper limit. The standard Triangle Inequality won't help us here because it provides a lower bound for the sum, not an upper bound.

We need the Reverse Triangle Inequality: $|a + b| \ge ||a| - |b||$. Let us regroup our original expression as $|(z^2 + z) + 1| = 1$. Applying the reverse inequality:

$$1 = | (z^2 + z) + 1 | \ge | | z^2 + z | - 1 |$$

This implies $|| z^2 + z | - 1 | \le 1$. If the absolute value of a quantity is less than or equal to $1$, then the quantity itself must lie between $-1$ and $1$. Thus, $-1 \le | z^2 + z | - 1 \le 1$.

Adding $1$ to all sides, we get $0 \le | z^2 + z | \le 2$. We have found that $| z^2 + z | \le 2$.

But we want $|z|$. We apply the reverse triangle inequality again on $| z^2 + z | = |z||z+1|$. Combining this with our upper bound of $2$, we get:

$$| |z|^2 - |z| | \le | z^2 + z | \le 2$$

Let $r = |z|$. We are solving $| r^2 - r | \le 2$, which leads to $r^2 - r - 2 \le 0$. Factoring this, we get $(r - 2)(r + 1) \le 0$.

Since $r$ is a modulus, $r \ge 0$, so $r+1$ is always positive. This forces $r - 2 \le 0$, or $r \le 2$. Thus, $|z| \le 2$. **Option B is confirmed as TRUE.**

### Phase 4: Debunking the Finite Set Myth

Finally, let us address the claim that $S$ has exactly four elements. As we discussed, the equation $| z^2 + z + 1 | = 1$ implies that $z^2 + z + 1$ lies on the unit circle.

We can write $z^2 + z + 1 = e^{i	heta}$ for any real $	heta$. This is a quadratic equation in $z$ for every single value of $	heta$.

Since $	heta$ is continuous and can take infinitely many values, there are infinitely many solutions for $z$. The set $S$ is not a collection of four points; it is a continuous curve. Therefore, the claim that $S$ has exactly four elements is **false**.

### Conclusion

We have navigated the geometry of the complex plane, used the Triangle and Reverse Triangle Inequalities to bound our variables, and debunked a common misconception about the nature of the solution set.

Remember, in JEE Advanced, the math is the language, but the geometry is the story. Keep visualizing, keep questioning, and keep pushing the boundaries of your understanding. You are doing great!

Animated Solution for Mathematics - Complex Numbers: Let $S$ be the set of all complex numbers $z$ satisfying $∣ z^{2} + z + 1∣ = 1$ . Then which of the following statements is/are TRUE?

Select Answer:

Visualized Solution

$S = {z \in C : ∣ z^{2} + z + 1∣ = 1}$

Completing the Square

Applying Triangle Inequality

Bounding the Modulus

First Conclusion

Regrouping for $∣ z ∣$

Reverse Triangle Inequality

Bounding $∣ z^{2} + z ∣$

Lower Bound of $∣ z^{2} + z ∣$

Solving for $r$

Second Conclusion

Number of Elements in $S$

Final Takeaways

The Sigma Insight: Algebraic Operations on Complex Numbers

The Geometry of Complex Numbers

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

Similar Questions

Let $S = {z = x + i y : \frac{2 z - 3 i}{4 z + 2 i} is a real number}$ . Then which of the following is NOT correct?

Let $z$ be a complex number such that the imaginary part of $z$ is non-zero and $a = z^{2} + z + 1$ is real. Then $a$ cannot take the value

The set of all $α \in R$ , for which $w = \frac{1 + ( 1 - 8 α ) z}{1 - z}$ is a purely imaginary number, for all $z \in C$ satisfying $∣ z ∣ = 1$ and $Re z \neq = 1$ , is :

If $z = \frac{1}{2} - 2 i$ , is such that $∣ z + 1∣ = α z + β (1 + i), i = - 1$ and $α, β \in R$ , then $α + β$ is equal to

Let $a, b, x$ and $y$ be real numbers such that $a - b = 1$ and $y \neq = 0$ . If the complex number $z = x + i y$ satisfies $Im (\frac{a z + b}{z + 1}) = y$ , then which of the following is(are) possible value(s) of $x$ ?

If $i z^{3} + z^{2} - z + i = 0$ , then show that $∣ z ∣ = 1$ .

Sum of squares of modulus of all the complex numbers $z$ satisfying $\overset{z}{ˉ} = i z^{2} + z^{2} - z$ is equal to \_\_\_\_\_.

Animated Solution for Mathematics - Complex Numbers: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let S be the set of all complex numbers z satisfying ∣z2+z+1∣=1. Then which of the following statements is/are TRUE?