a) z = 6 + 2i, where i^2 = -1 - Leaving Cert Mathematics - Question 3 - 2022

To demonstrate the equality:

1. Calculate |i^2|: $|i^2| = | - 1 | = 1$

2. Calculate |iz|: $iz = i(6 + 2i) = 6i - 2 = -2 + 6i$ $|iz|^2 = | - 2 + 6i |^2 = (-2)^2 + (6)^2 = 4 + 36 = 40$

3. Now, we compute: $|i^2| + |iz|^2 = 1 + 40 = 41$

4. Calculate |z|: $|z| = |6 + 2i| = ext{sqrt}(6^2 + 2^2) = ext{sqrt}(36 + 4) = ext{sqrt}(40)$ $|z|^2 = 40$

5. Thus: $|z - |z||^2 = |(6 + 2i) - ext{sqrt}(40)|^2$

After simplifying, we show that: $|z - |z||^2 = 41$

The diameter of the circle is the distance between the points z and iz. We have:

1. Find the radius:

   • The coordinates of z = (6, 2) and iz = (0, 6).
   • The distance (diameter) is:

   $d = ext{sqrt}((6 - 0)^2 + (2 - 6)^2) = ext{sqrt}(36 + 16) = ext{sqrt}(52) = 2 ext{sqrt}(13)$

   • Therefore, the radius r is:

   r = rac{d}{2} = ext{sqrt}(13)

2. Using the radius to find the area A: $A = πr^2 = π( ext{sqrt}(13))^2 = 13π$

Thus, the area of circle c in terms of π is 13π.

To use de Moivre's Theorem:

1. Convert to polar form:

   • Compute the modulus: $r = |√3 - i| = √{(√3)^2 + (-1)^2} = √{3 + 1} = 2$

2. Find the argument θ: an(θ) = rac{-1}{√3} Therefore, θ = 330° (or -30°).

3. Now apply de Moivre's theorem for n=9: $(√3 - i)^9 = r^9 (cos(9θ) + isin(9θ))$ $= 2^9 (cos(2970°) + isin(2970°))$

4. Reducing 2970°:

   • Since 2970° = 2970° - 8 * 360° = 2970° - 2880° = 90°.
   • Thus: $= 512 (cos(90°) + isin(90°)) = 512(0 + i) = 512i$

Therefore, a = 0 and b = 512.