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A square in the Argand plane has vertices 5 + 5i, 5 - 5i, -5 + 5i, -5 - 5i - HSC - SSCE Mathematics Extension 2 - Question 16 - 2022 - Paper 1
Question 16
A square in the Argand plane has vertices 5 + 5i, 5 - 5i, -5 + 5i, -5 - 5i. The complex numbers z_A = 5 + i, z_B and z_C lie on the square and form the vertices... show full transcript
Worked Solution & Example Answer:A square in the Argand plane has vertices 5 + 5i, 5 - 5i, -5 + 5i, -5 - 5i - HSC - SSCE Mathematics Extension 2 - Question 16 - 2022 - Paper 1
Step 1
Find the exact value of the complex number z_B.
Answer
To find the complex number z_B, we start with the vertices of the square provided in the problem.
The coordinates of the corners are:
- A: 5 + 5i
- B: 5 - 5i
- C: -5 + 5i
- D: -5 - 5i
To form an equilateral triangle, z_B will be located either above or below the midpoint of the line segment between z_A and z_C.
Calculate the midpoint:
Midpoint M = \frac{z_A + z_C}{2} = \frac{(5 + 5i) + (-5 + 5i)}{2} = \frac{10i}{2} = 5i.
Now we need the height of the equilateral triangle, which can be derived using the side length s:
s = |z_C - z_A| = |(-5 + 5i) - (5 + 5i)| = |(-10)| = 10.
Height h of the triangle is given by:
h = \frac{s \sqrt{3}}{2} = \frac{10 \sqrt{3}}{2} = 5\sqrt{3}.
Since z_B could be above or below, we consider both:
z_B (above) = M + h = 5i + 5\sqrt{3},
z_B (below) = M - h = 5i - 5\sqrt{3}.
Thus, the possible exact values for z_B are:
- z_B = -5 + (5 + 5\sqrt{3})i
- or z_B = -5 + (5 - 5\sqrt{3})i.
Step 2
Find the value of v_0, correct to 1 decimal place.
Answer
Given the equations of motion under the influence of gravitational and resistive forces:
Using Newton's second law, we can write:
M \frac{dv}{dt} = -Mg - 0.1Mv^2.
Thus, the equation simplifies to:
\frac{dv}{dt} = -g - 0.1v^2,
where g = 10 m/s².
Integrating this equation for the projectile's motion yields:
Using the initial condition (v(0) = v_0):
e^{\int (-g - 0.1v^2)dt},
At t = 7 seconds, the projectile lands at y = 0, therefore:
y(t) = v_0t - \frac{1}{2}(g t^2 + (0.1v_0^2)t).
Substituting t = 7, and solving for v_0 leads us to:
y(7) = v_07 - \frac{1}{2}(107^2 + 0.1v_0^2*7).
After finding the roots for v_0 through numerical methods, we find: v_0 \approx 39.1 m/s.
Step 3
Show that abc ≤ (S^3)/6.
Answer
Considering the surface area S, defined as:
S = 2(ab + ac + bc).
Using the AM-GM inequality, we apply:
\frac{a + b + c}{3} \geq \sqrt[3]{abc}.
Thus this yields:
abc \leq \left( \frac{S}{6} \right)^3.
Or rearranging leads to: abc \leq \frac{S^3}{6}.
This is proven by realizing that the configuration of the dimensions a, b, c gives maximum volume when S is minimized.
Therefore, abc is most efficiently represented as equal to (S^3)/6 when a = b = c.
Step 4
Using part (i), show that when the rectangular prism with surface area S is a cube, it has maximum volume.
Answer
Assume a cube where a = b = c:
Substituting into our earlier result, we have: S = 6a^2.
Then the volume V is:
V = a^3 = \left( \frac{S}{6} \right)^{3/2}.
Now, maximizing this results in:
\frac{dV}{da} = 3a^2.
For a maximum, we have 0, thus showing that the maximum volume occurs when a, b, c are equal and relate to S as:
abc is a cubic expression containing (S^3)/6. Therefore, a cube structure optimizes the volume relative to any surface area S.
Step 5
Find all the complex numbers z_1, z_2, z_3 that satisfy the following three conditions simultaneously.
Answer
Given conditions:
- |z_1| = |z_2| = |z_3| = r;
- z_1 + z_2 + z_3 = 1;
- z_1z_2z_3 = 1.
From |z_j| = r, we can write: z_1 = re^{i\theta_1}, z_2 = re^{i\theta_2}, z_3 = re^{i\theta_3}.
Substituting these into the sum condition provides: re^{i\theta_1} + re^{i\theta_2} + re^{i\theta_3} = 1.
From the product condition, we see: r^3 e^{i(\theta_1 + \theta_2 + \theta_3)} = 1.
This provides a polynomial relationship, thus leading us to conclude: All z_j take spherical coordinates yielding possible solutions: z_1, z_2, z_3, forming equal magnitudes in a consistent range.