Key Concepts and Formulas

• Modulus of a Complex Number: For $z = a+ib$, the modulus is $|z| = \sqrt{a^2+b^2}$. Key properties include $|z^k| = |z|^k$ and $|\bar{z}| = |z|$. Also, $|zw| = |z||w|$.
• Properties of Conjugates: For $z = a+ib$, its conjugate is $\bar{z} = a-ib$. If $|z|=1$, then $z\bar{z} = |z|^2 = 1$, which implies $\bar{z} = 1/z$.
• Cube Roots of Unity: The solutions to $z^3=1$ are $1, \omega, \omega^2$, where $\omega = e^{i2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}$ and $\omega^2 = e^{i4\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$. They satisfy $1+\omega+\omega^2=0$ and $\omega^3=1$.

Step-by-Step Solution

Step 1: Find the modulus of z

We are given the equation $z^2 = \bar{z} \cdot 2^{1-|z|} \quad (*)$ To find $|z|$, we take the modulus of both sides of the equation: $|z^2| = |\bar{z} \cdot 2^{1-|z|}|$ Using the properties of the modulus, we have $|z^2| = |z|^2$, $|\bar{z}| = |z|$, and since $2^{1-|z|}$ is a positive real number, $|2^{1-|z|}| = 2^{1-|z|}$. Therefore, $|z|^2 = |\bar{z}| \cdot |2^{1-|z|}| = |z| \cdot 2^{1-|z|}$ Since $z = a+ib$ with $b \neq 0$, we know that $z \neq 0$, so $|z| \neq 0$. We can thus divide both sides by $|z|$: $|z| = 2^{1-|z|}$ Let $k = |z|$. We need to solve $k = 2^{1-k}$. By inspection, $k=1$ is a solution since $1 = 2^{1-1} = 2^0 = 1$. To show uniqueness, let $f(k) = k - 2^{1-k}$. Then $f'(k) = 1 + (\ln 2) 2^{1-k} > 0$ for all $k$. Thus $f(k)$ is strictly increasing, and $k=1$ is the only solution. Therefore, we conclude that $|z| = 1$

Step 2: Determine the value of z

Substituting $|z|=1$ back into the original equation $(*)$, we get: $z^2 = \bar{z} \cdot 2^{1-1} = \bar{z} \cdot 2^0 = \bar{z}$ Since $|z|=1$, we know that $\bar{z} = \frac{1}{z}$. Substituting this into the equation, we get: $z^2 = \frac{1}{z}$ $z^3 = 1$ The solutions to $z^3=1$ are the cube roots of unity: $z = 1, \quad z = e^{i2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}, \quad z = e^{i4\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$ Since the problem states that $b \neq 0$, $z$ must have a non-zero imaginary part. This excludes $z=1$. Therefore, $z$ must be one of the two non-real cube roots of unity, commonly denoted as $\omega$ and $\omega^2$: $z = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \quad \text{or} \quad z = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$

Step 3: Find the least value of n

We are looking for the least value of $n \in \mathbb{N}$ such that $z^n = (z+1)^n$. Since $z \neq 0$, we can divide by $z^n$: $\left(\frac{z+1}{z}\right)^n = 1$ This can be rewritten as: $\left(1 + \frac{1}{z}\right)^n = 1$ We consider the two possible values for $z$:

Case 1: $z = -\frac{1}{2} + i\frac{\sqrt{3}}{2} = \omega$ Since $|z|=1$, we have $1/z = \bar{z} = \bar{\omega} = \omega^2$. $1 + \frac{1}{z} = 1 + \omega^2 = -\omega$ Substituting this into our equation: $(-\omega)^n = 1$ $(-1)^n \omega^n = 1$ If $n$ is odd, then $-\omega^n = 1$, so $\omega^n = -1$. Since $\omega^3=1$, the powers of $\omega$ are $1, \omega, \omega^2$. None of these equals $-1$, so $n$ cannot be odd. If $n$ is even, then $\omega^n = 1$. Since $\omega^3=1$, we must have $n$ be a multiple of 3. Thus, $n$ must be an even multiple of 3. The smallest such $n$ is $n=6$.

Case 2: $z = -\frac{1}{2} - i\frac{\sqrt{3}}{2} = \omega^2$ Since $|z|=1$, we have $1/z = \bar{z} = \overline{\omega^2} = \omega$. $1 + \frac{1}{z} = 1 + \omega = -\omega^2$ Substituting this into our equation: $(-\omega^2)^n = 1$ $(-1)^n (\omega^2)^n = 1$ If $n$ is odd, then $-(\omega^2)^n = 1$, so $(\omega^2)^n = -1$. Since $(\omega^2)^3=1$, the powers of $\omega^2$ are $1, \omega^2, \omega$. None of these equals $-1$, so $n$ cannot be odd. If $n$ is even, then $(\omega^2)^n = 1$. Since $(\omega^2)^3=1$, we must have $n$ be a multiple of 3. Thus, $n$ must be an even multiple of 3. The smallest such $n$ is $n=6$.

In both possible cases for $z$, the least value of $n \in \mathbb{N}$ for which the condition holds is 6.

Common Mistakes & Tips

• Forgetting to apply all given conditions, such as $b \neq 0$, which would lead to including $z=1$ as a possible value.
• When solving equations like $w^n=1$, remember that $n$ must be a multiple of the order of $w$. The order of $w$ is the smallest positive integer $k$ such that $w^k=1$. For primitive cube roots of unity, the order is 3.
• When an equation involves both $z$ and $\bar{z}$, or powers of $z$ and a constant base raised to a power involving $|z|$, taking the modulus of both sides is often the first and most effective step to simplify the equation and find $|z|$.

Summary

The problem requires a systematic approach, starting by determining the magnitude of the complex number $z$, then its specific value(s), and finally solving for $n$. By taking the modulus of the initial equation, we found that $|z|=1$. Substituting $|z|=1$ back into the original equation simplified it to $z^2=\bar{z}$, which implies $z^3=1$. The condition $b \neq 0$ was crucial to select the correct non-real cube roots of unity for $z$. The second condition $z^n=(z+1)^n$ was transformed into $(1+1/z)^n=1$, leading to $n=6$.

Final Answer

The final answer is $\boxed{6}$.