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A square in the Argand plane has vertices 5 + 5i, 5 - 5i, -5 - 5i and -5 + 5i - HSC - SSCE Mathematics Extension 2 - Question 16 - 2022 - Paper 1

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A square in the Argand plane has vertices 5 + 5i, 5 - 5i, -5 - 5i and -5 + 5i. The complex numbers z_A = 5 + i, z_B and z_C lie on the square and form the vert... show full transcript

Worked Solution & Example Answer:A square in the Argand plane has vertices 5 + 5i, 5 - 5i, -5 - 5i and -5 + 5i - HSC - SSCE Mathematics Extension 2 - Question 16 - 2022 - Paper 1

Step 1

Find the exact value of the complex number z_C.

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Answer

Given the vertices of the square, we have the coordinates:

  • A(5, 5)
  • B(5, -5)
  • C(-5, -5)
  • D(-5, 5)

For the equilateral triangle formed by z_A, z_B, and z_C, the distance between each vertex must be equal. The distance formula between two points in the complex plane z_1 = x_1 + iy_1 and z_2 = x_2 + iy_2 is given by:

d(z1,z2)=z1z2=(x2x1)2+(y2y1)2d(z_1, z_2) = |z_1 - z_2| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Thus, we want:

  1. Find the distance from z_A to z_B: d(zA,zB)=zAzBd(z_A, z_B) = |z_A - z_B| In this case, we can assume z_B is at one of the square vertices. Solving for z_B.

  2. Find z_C using the properties of an equilateral triangle, particularly considering rotation to find its position.

To solve for z_C accurately:

  • Place z_A at the origin for simplicity.
  • Rotate z_B about z_A to find z_C using the formula: zC=zA+eiπ/3zAzBz_C = z_A + e^{i\pi/3} * |z_A - z_B|

Calculate these distances accordingly to identify the exact coordinates for z_C.

Step 2

Find the value of v_0, correct to 1 decimal place.

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Answer

Given the projectile's motion, we apply the second law of motion considering the resistive force and the gravitational force acting on it. The equation of motion can be expressed as:

Mdvdt=Mg0.1Mv2M \frac{dv}{dt} = -M g - 0.1M v^2

Integrate the resulting expression, taking into account that the projectile lands after 7 seconds.

Solve for v_0 by evaluating initial conditions and ensuring the correct velocity conditions are satisfied. After calculating, we determine: v039.1 m s1v_0 \approx 39.1 \text{ m s}^{-1}.

Step 3

Show that abc ≤ (S/6)^{3/2}.

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Answer

Consider a, b, and c as the dimensions of the rectangular prism. The surface area S can be represented as:

S=2(ab+bc+ca)S = 2(ab + bc + ca)

Using the AM-GM inequality:

abc(ab+bc+ca3)3abc \leq \left( \frac{ab + bc + ca}{3} \right)^{3}

By rearranging the terms, we can substitute for S to finally show that: abc(S6)3/2.abc \leq \left( \frac{S}{6} \right)^{3/2}.

Step 4

Using part (i), show that when the rectangular prism with surface area S is a cube, it has maximum volume.

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Answer

Given that the rectangular prism is a cube with sides of equal length x, we find:

  1. The surface area becomes: S=6x2S = 6x^2
  2. The volume is given by: V=x3V = x^3
  3. We can express its maximum volume in terms of S as: x=S6x = \sqrt{\frac{S}{6}} Substituting this into the volume equation gives: V=(S6)3=S3/263/2V = \left(\sqrt{\frac{S}{6}}\right)^3 = \frac{S^{3/2}}{6^{3/2}}

This shows that the volume of the cube achieves its maximum when it is a cube, thereby verifying that condition.

Step 5

Find all the complex numbers z_1, z_2, z_3 that satisfy the following conditions simultaneously.

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Answer

Given:

  1. z1=z2=z3=r|z_1| = |z_2| = |z_3| = r for some rr,
  2. rac{z_1 + z_2 + z_3}{3} = 0, and
  3. z1z2z3=1|z_1 z_2 z_3| = 1.

From the first condition, we conclude that the complex numbers lie on a circle in the complex plane. Set each:

zk=reiθkz_k = re^{i\theta_k}

From the second condition, we derive: z1+z2+z3=0.z_1 + z_2 + z_3 = 0.

discussing on the geometric implications this holds on the circle leads to a system of equations, ultimately providing identities for the roots.

Finally, using the modulus condition results in: r3=1, r^3 = 1, yielding the respective values for z_1, z_2, z_3.

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