Find, in a simplified form, the cartesian equation of the locus of the point P representing the complex number z where $|z+3|=2\,\Re(z)$. Sketch the locus on an Argand diagram. Write down the modulus and argument of the complex number $w$, where $w=-2+(2\sqrt{3})\,i$. Given that $w^{n}$ is real, find the possible values of $n$, where $n$ is a positive integer.
Express $w^{50}-(w^{*})^{50}$ in the form $k\,i$, where $k$ is real, giving the exact value of $k$ in non-trigonometrical form.

1. Substitute $z = x + iy$ and apply the modulus and real part definitions.

Let $z = x + iy$.
$|x + iy + 3| = 2\Re(x + iy)$
$|x + 3 + iy| = 2x$
$\sqrt{(x+3)^2 + y^2} = 2x$

2. Establish the domain restriction before squaring to avoid extraneous solutions.

Since the modulus of a complex number is always non-negative, the right hand side must also be non-negative:
$2x \ge 0 \implies x \ge 0$

3. Square both sides and simplify the algebraic equation.

$(x+3)^2 + y^2 = (2x)^2$
$x^2 + 6x + 9 + y^2 = 4x^2$
$3x^2 - 6x - y^2 = 9$

4. Complete the square to express the locus in standard form and finalise the domain.

$3(x^2 - 2x) - y^2 = 9$
$3[(x-1)^2 - 1] - y^2 = 9$
$3(x-1)^2 - y^2 = 12$
$\frac{(x-1)^2}{4} - \frac{y^2}{12} = 1$

Given the initial condition $x \ge 0$, we observe that $\frac{(x-1)^2}{4} = 1 + \frac{y^2}{12} \ge 1$.
This implies $(x-1)^2 \ge 4$, so $x - 1 \ge 2$ or $x - 1 \le -2$.
Since $x \ge 0$, we must have $x \ge 3$.
Thus, the locus is the right branch of the hyperbola $\frac{(x-1)^2}{4} - \frac{y^2}{12} = 1$.

5. Describe the correct sketch on the Argand diagram.

The sketch should consist of:

• A set of real (Re) and imaginary (Im) axes.
• The point $(1,0)$ indicating the intersection of asymptotes.
• The asymptotes $y = \pm\sqrt{3}(x-1)$ drawn as dotted lines.
• A single continuous curve (the right branch of the hyperbola) passing through the vertex at $(3,0)$ and approaching the asymptotes as $x \to \infty$.

1. Find the modulus and argument of $w$.

$|w| = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = 4$

Since $w$ is in the second quadrant, its argument is:
$\arg(w) = \pi - \tan^{-1}\left(\frac{2\sqrt{3}}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$

2. Express $w^n$ in polar form and set the condition for it to be real.

$w = 4\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)$

Using de Moivre's theorem:
$w^n = 4^n\left(\cos\frac{2n\pi}{3} + i\sin\frac{2n\pi}{3}\right)$

For $w^n$ to be real, the imaginary part must be zero:
$\sin\frac{2n\pi}{3} = 0$

3. Solve for the possible values of $n$.

$\frac{2n\pi}{3} = m\pi$, where $m \in \mathbb{Z}$
$n = \frac{3m}{2}$

Since $n$ is a positive integer ($n \in \mathbb{Z}^+$), $m$ must be an even positive integer (let $m = 2p$).
$n = 3p$
Thus, the possible values of $n$ are all positive multiples of $3$.

4. Express $w^{50} - (w^*)^{50}$ using polar form.

Using the properties of the conjugate, $w^* = 4\left(\cos\left(-\frac{2\pi}{3}\right) + i\sin\left(-\frac{2\pi}{3}\right)\right)$

$w^{50} - (w^*)^{50} = 4^{50}\left(\cos\frac{100\pi}{3} + i\sin\frac{100\pi}{3}\right) - 4^{50}\left(\cos\left(-\frac{100\pi}{3}\right) + i\sin\left(-\frac{100\pi}{3}\right)\right)$

Since $\cos(-\theta) = \cos\theta$ and $\sin(-\theta) = -\sin\theta$, this simplifies to:
$w^{50} - (w^*)^{50} = 4^{50}\left(2i\sin\frac{100\pi}{3}\right)$

5. Evaluate the trigonometric term to find the exact value of $k$.

Simplify the angle:
$\frac{100\pi}{3} = 33\pi + \frac{\pi}{3} = 16(2\pi) + \pi + \frac{\pi}{3}$
$\sin\left(16(2\pi) + \pi + \frac{\pi}{3}\right) = \sin\left(\pi + \frac{\pi}{3}\right) = -\sin\frac{\pi}{3} = -\frac{\sqrt{3}}{2}$

Substitute back into the expression:
$w^{50} - (w^*)^{50} = 4^{50}\left(2i\left(-\frac{\sqrt{3}}{2}\right)\right) = -4^{50}\sqrt{3}\,i$

Recognising $4^{50} = (2^2)^{50} = 2^{100}$, we have $-2^{100}\sqrt{3}\,i$.
Therefore, $k = -2^{100}\sqrt{3}$.