I can see the examination card with multiple math problems. Let's work through them one by one. I'll provide the steps for each solution:

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1. Evaluate $z = (\sqrt{3} + i)^6$

To evaluate this, we can convert the complex number $z = \sqrt{3} + i$ to polar form.

• The modulus of $z$ is: $|z| = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$
• The argument of $z$, $\theta$, is: $\theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$

Now, express $z$ in polar form: $z = 2\left(\cos\frac{\pi}{6} + i \sin\frac{\pi}{6}\right)$

To evaluate $z^6$, use De Moivre’s Theorem: $z^6 = 2^6 \left( \cos\left(6 \times \frac{\pi}{6}\right) + i \sin\left(6 \times \frac{\pi}{6}\right) \right) = 64 \left( \cos \pi + i \sin \pi \right)$ Since $\cos \pi = -1$ and $\sin \pi = 0$, we get: $z^6 = 64 \times (-1) = -64$

2. Graphically represent the set of values of $z$ for which $1 \leq |z| \leq 3$, Im $z \geq 1$

This problem describes a region in the complex plane:

• The modulus of $z$, $|z|$, must lie between 1 and 3, which corresponds to a circular annular region with inner radius 1 and outer radius 3.
• The imaginary part of $z$, $\text{Im}(z)$, must be greater than or equal to 1, meaning the region is above the line $\text{Im}(z) = 1$.

This can be visualized as a ring between two circles (radius 1 and radius 3), limited to the upper half-plane above $\text{Im}(z) = 1$.

3. Solve the equation $z^2 - i = 0$

Rearrange the equation: $z^2 = i$

To solve for $z$, express $i$ in polar form: $i = e^{i\frac{\pi}{2}}$

Thus, $z^2 = e^{i\frac{\pi}{2}}$, and the general solution for $z$ is: $z = e^{i\frac{\pi}{4} + i\frac{n\pi}{2}} \quad \text{for} \quad n = 0, 1$

The two solutions are: $z = e^{i\frac{\pi}{4}} \quad \text{or} \quad z = e^{i\frac{5\pi}{4}}$ In rectangular form: $z = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \quad \text{or} \quad z = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}$

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Would you like me to continue solving the rest of the problems?