Let's tackle each part of the problem step by step.

(ii) $Z_2 - Z_1^2$

Given:

• $Z_1 = 3 + 2i$
• $Z_2 = 13 - i$

First, calculate $Z_1^2$:

$Z_1^2 = (3 + 2i)^2 = 9 + 12i + 4i^2$ Since $i^2 = -1$:

$Z_1^2 = 9 + 12i - 4 = 5 + 12i$

Now, subtract $Z_1^2$ from $Z_2$:

$Z_2 - Z_1^2 = (13 - i) - (5 + 12i) = 13 - 5 - i - 12i = 8 - 13i$

So, $Z_2 - Z_1^2 = 8 - 13i$.

(b) Represent $2 = (\cos \theta + i \sin \theta)$ in algebraic form

This represents a complex number in polar form. In algebraic (rectangular) form, it's:

$Z = 2 (\cos \theta + i \sin \theta) = 2\cos \theta + 2i\sin \theta$

(c) $Z_1 = [3, 520^\circ]$ and $Z_2 = [2, 15^\circ]$, find $Z_1Z_2$

First, convert the angles to a standard range (if necessary), but in this case:

• $Z_1 = 3 \text{cis}(520^\circ)$
• $Z_2 = 2 \text{cis}(15^\circ)$

The product $Z_1Z_2$ in polar form:

$Z_1Z_2 = (3 \times 2) \text{cis}(520^\circ + 15^\circ) = 6 \text{cis}(535^\circ)$

Since $535^\circ = 535^\circ - 360^\circ = 175^\circ$:

$Z_1Z_2 = 6 \text{cis}(175^\circ)$

(d) Express $Z = -2 + 2\sqrt{3}i$ in trigonometric form

First, find the modulus $|Z|$:

$|Z| = \sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = \sqrt{16} = 4$

Now, find the argument $\theta$ using $\tan \theta = \frac{2\sqrt{3}}{-2} = -\sqrt{3}$:

Since the complex number is in the second quadrant (negative real part and positive imaginary part), $\theta = 150^\circ$ or $\theta = \frac{5\pi}{6}$ radians.

So the trigonometric form is:

$Z = 4\left(\cos 150^\circ + i \sin 150^\circ\right)$

(e) Find the modulus and argument of the complex number

To solve this problem, we can either express the complex number or use the modulus and argument formulas. However, there is no specific complex number given, so please provide the complex number to proceed.

(f) Given $Z = a + ib$, find $a$ and $b$ by solving the simultaneous equations

• $ZZ^* = 65$
• $Z - 2 = 16$

Let $Z = a + ib$ and $Z^* = a - ib$, then:

$ZZ^* = (a + ib)(a - ib) = a^2 + b^2 = 65$

And:

$Z - 2 = a + ib - 2 = 16 \quad \text{(This equation seems unusual; please check if it's correct.)}$

If the second equation is correct as $a + ib = 18$, then $a = 18$ and $b = 0$, which conflicts with the first equation. We might need to clarify this step or recheck the problem.

(g) Show that for the complex variable $Z$, $\cos^2 \theta + \sin^2 \theta = 1$

This is a trigonometric identity for any angle $\theta$, which holds for all complex numbers. If $Z = r(\cos \theta + i \sin \theta)$, then:

$\cos^2 \theta + \sin^2 \theta = 1$

This is a standard trigonometric identity derived from the Pythagorean theorem.

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Do you need further explanations on any specific part of the problem?

Here are five related questions to deepen your understanding:

1. How do you convert a complex number from rectangular form to polar form?
2. What is the geometric interpretation of the modulus of a complex number?
3. How does multiplying two complex numbers in polar form affect their arguments and moduli?
4. Can you solve the simultaneous equations $Z^2 = 65$ and $Z = 16$?
5. What are the common mistakes to avoid when working with complex numbers?

Tip: When solving complex numbers in polar form, always ensure the argument is within the standard range (usually $-180^\circ$ to $180^\circ$).