Let z = a + ib, b ≠ 0 be complex numbers satisfying z² = z̄ · 2^(1-|z|). Then the least value of n ∈

Q: Why can we just take the modulus on both sides? Doesn't that lose information about the argument of $z$?

That is a brilliant question! Taking the modulus does indeed discard the argument, but it is a strategic move. We are not trying to solve for $z$ immediately; we are trying to find its magnitude. Once we know $|z|=1$, we can substitute it back into the original equation to find the argument. It is a two-step dance, not a loss of information.

Q: Why is $z=1$ rejected?

The problem explicitly states $b \neq 0$, where $z = a + ib$. This means $z$ cannot be a purely real number. Since $z=1$ is a real number, it violates the given constraint. We must always keep an eye on these subtle constraints; they are often the difference between a correct answer and a trap.

Q: How do I know $z^3=1$?

Great observation! We found $|z|=1$, which implies $z \bar{z} = 1$, or $\bar{z} = \frac{1}{z}$. Substituting this into the simplified equation $z^2 = \bar{z}$ gives $z^2 = \frac{1}{z}$, which leads directly to $z^3 = 1$. This is the beauty of complex numbers—they often collapse into these elegant, symmetric forms.

Question

Let $z = a + ib, b 
eq 0$ be complex numbers satisfying $z^2 = \bar{z} \cdot 2^{1-|z|}$. Then the least value of $n \in \mathbb{N}$, such that $z^n = (z + 1)^n$, is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the complex plane, looking at the equation $z^2 = \bar{z} \cdot 2^{1-|z|}$. In the world of JEE Advanced, chaos is just order waiting to be discovered.

We are given $z = a + ib$ with $b 
eq 0$, which tells us immediately that $z$ is not a real number. It is a point dancing somewhere off the real axis.

To tame this beast, we take the modulus of both sides:
$$|z^2| = |\bar{z}| \cdot |2^{1-|z|}|$$

Since $|z^2| = |z|^2$ and $|\bar{z}| = |z|$, and $2^{1-|z|}$ is always positive, the equation simplifies to:
$$|z|^2 = |z| \cdot 2^{1-|z|}$$

Since $z 
eq 0$, we can divide by $|z|$ to obtain:
$$|z| = 2^{1-|z|}$$

If we let $|z| = r$, we look for the intersection of the line $f(r) = r$ and the exponential decay $g(r) = 2^{1-r}$. By inspection, $r=1$ is the unique solution. Our complex number $z$ is trapped on the unit circle.

### The Elegance of Roots of Unity

Now that we know $|z|=1$, the term $2^{1-|z|}$ becomes $2^0$, which is $1$. The original equation collapses to:
$$z^2 = \bar{z}$$

On the unit circle, we know that $z \bar{z} = |z|^2 = 1$, which implies $\bar{z} = \frac{1}{z}$. Substituting this into our simplified equation, we get:
$$z^2 = \frac{1}{z} \Rightarrow z^3 = 1$$

This is the famous equation for the cube roots of unity. The solutions are $1, \omega, 	ext{ and } \omega^2$.

However, recall our constraint $b 
eq 0$. This means $z$ cannot be real, so $z$ cannot be $1$. Thus, $z$ must be either $\omega$ or $\omega^2$.

### The Final Challenge

We are tasked with finding the least natural number $n$ such that $z^n = (z+1)^n$. Substituting $z = \omega$, we get:
$$\omega^n = (\omega+1)^n$$

Recall the fundamental identity of cube roots of unity: $1 + \omega + \omega^2 = 0$. This implies $\omega + 1 = -\omega^2$. Our equation becomes:
$$\omega^n = (-\omega^2)^n$$

This simplifies to:
$$\omega^n = (-1)^n \cdot \omega^{2n}$$

Dividing both sides by $\omega^n$, we obtain:
$$1 = (-1)^n \cdot \omega^n \Rightarrow (-\omega)^n = 1$$

Now, we test values of $n$ to find the smallest natural number:
- For $n=1$, $-\omega 
eq 1$.
- For $n=2$, $\omega^2 
eq 1$.
- For $n=3$, $-\omega^3 = -1 
eq 1$.
- For $n=4$, $\omega^4 = \omega 
eq 1$.
- For $n=5$, $-\omega^5 = -\omega^2 
eq 1$.
- For $n=6$, $(-\omega)^6 = (-1)^6 \cdot \omega^6 = 1 \cdot 1 = 1$.

The least natural number $n$ is **6**.

Animated Solution for Mathematics - Complex Numbers: Let $z = a + ib, b \neq = 0$ be complex numbers satisfying $z^{2} = \overset{z}{ˉ} \cdot 2^{1 - ∣ z ∣}$ . Then the least value of $n \in N$ , such that $z^{n} = (z + 1)^{n}$ , is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Complex Equation

Apply Modulus Property

Solve for $∣ z ∣$

Simplify the Original Equation

Relate to Cube Roots of Unity

Identify the Value of $z$

Set up the Target Equation

Apply the Identity $1 + ω + ω^{2} = 0$

Simplify the Power Equation

Test Values for $n$

Conclusion and Final Answer

The Sigma Insight: Cube Roots and nth Roots of Unity

Analyzing the Setup

The Elegance of Roots of Unity

The Final Challenge

Similar Questions

If $z^{2} + z + 1 = 0$ , where $z$ is complex number, then the value of $(z + 1/ z)^{2} + (z^{2} + 1/ z^{2})^{2} + (z^{3} + 1/ z^{3})^{2} + \dots + (z^{6} + 1/ z^{6})^{2}$ is

If $ω (\neq = 1)$ be a cube root of unity and $(1 + ω^{2})^{n} = (1 + ω^{4})^{n}$ , then the least positive value of $n$ is

If $z^{2} + z + 1 = 0, z \in C$ , then $\sum_{n = 1}^{15} (z^{n} + (- 1)^{n} \frac{1}{z ^{n}})^{2}$ is equal to \_\_\_\_\_.

Let $ω \neq = 1$ be a cube root of unity. Then the minimum of the set ${∣ a + bω + c ω^{2} ∣^{2} : a, b, c distinct non-zero integers}$ equals

Let $ω$ be a complex number such that $2 ω + 1 = z$ where $z = - 3$ . If $111 1 - ω^{2} - 1 ω^{2} 1 ω^{2} ω^{7} = 3 k$ , then k is equal to:

Let $z_{1}$ and $z_{2}$ be $n^{t h}$ roots of unity which subtend a right angle at the origin. Then $n$ must be of the form

Let a complex number $α, α \neq = 1$ , be a root of the equation $z^{p + q} - z^{p} - z^{q} + 1 = 0$ , where $p, q$ are distinct primes. Show that either $1 + α + α^{2} + \dots + α^{p - 1} = 0$ or $1 + α + α^{2} + \dots + α^{q - 1} = 0$ , but not both together.

The value of the expression $1 \cdot (2 - ω) (2 - ω^{2}) + 2 \cdot (3 - ω) (3 - ω^{2}) + \dots + (n - 1) \cdot (n - ω) (n - ω^{2})$ , where $ω$ is an imaginary cube root of unity, is $\dots$

If $α$ is a root of the equation $x^{2} + x + 1 = 0$ and $\sum_{k = 1}^{n} (α^{k} + \frac{1}{α ^{k}})^{2} = 20$ , then $n$ is equal to

If $z$ is a complex number, then the number of common roots of the equation $z^{1985} + z^{100} + 1 = 0$ and $z^{3} + 2 z^{2} + 2 z + 1 = 0$ , is equal to :

Enter Numerical Value:

Visualized Solution

Analyze the Complex Equation

Apply Modulus Property

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣z∣

Simplify the Original Equation

Relate to Cube Roots of Unity

Identify the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } z

Set up the Target Equation

Apply the Identity .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 1+ω+ω2=0

Simplify the Power Equation

Test Values for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Conclusion and Final Answer

The Sigma Insight: Cube Roots and nth Roots of Unity

Analyzing the Setup

The Elegance of Roots of Unity

The Final Challenge

Similar Questions

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