Solve the equation iz5=−32i z^5 = -32, giving your roots in the form reiθre^{i\theta}, where r>0r>0 and −π<θ≤π-\pi<\theta\le\pi. Sketch on an Argand diagram the points P1,P2,P3,P4P_1, P_2, P_3, P_4 and P5P_5 representing these roots, where P1P_1 represents the root with the smallest argument and P1P2P3P4P5P_1 P_2 P_3 P_4 P_5 is a polygon described in an anticlockwise sense. Find the area of P1P2P3P4P5P_1 P_2 P_3 P_4 P_5. On an Argand diagram sketch clearly the locus of PP where PP represents the complex number zz such that ∣z−2−2i∣≤1|z-2-2i|\le 1 and arg⁡(z−1)=arg⁡(1+3i)\arg(z-1)=\arg(1+3i). Find the range of values of arg⁡(z−3−2i)\arg(z-3-2i), given that ∣z−2−2i∣≤1|z-2-2i|\le 1 and arg⁡(z−1)=arg⁡(1+3i).\arg(z-1)=\arg(1+3i).

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Worked solution

8(a)

1. Rearrange the given equation to make $z^5$ the subject and express the right-hand side in exponential form.

iz5=−32i z^5 = -32
z5=−32i=−32(−i)=32iz^5 = \frac{-32}{i} = -32(-i) = 32i
To express 32i32i in exponential form reiθre^{i\theta}:
The modulus is 3232, and the argument is π2\frac{\pi}{2}.
z5=32ei(π2+2kπ) for k∈Zz^5 = 32e^{i\left(\frac{\pi}{2} + 2k\pi\right)} \text{ for } k \in \mathbb{Z}

2. Find the general expression for the roots by taking the 5th root.

z=(32)1/5ei(π2+2kπ5)z = (32)^{1/5} e^{i\left(\frac{\frac{\pi}{2} + 2k\pi}{5}\right)}
z=2ei(π10+2kπ5)z = 2 e^{i\left(\frac{\pi}{10} + \frac{2k\pi}{5}\right)}

3. Substitute integer values for $k$ to find the 5 roots within the principal argument range $-\pi < \theta \le \pi$.

Testing consecutive integers for kk:
For k=0k = 0, θ=π10  ⟹  z=2eiπ10\theta = \frac{\pi}{10} \implies z = 2e^{i\frac{\pi}{10}}
For k=1k = 1, θ=π10+4π10=π2  ⟹  z=2eiπ2\theta = \frac{\pi}{10} + \frac{4\pi}{10} = \frac{\pi}{2} \implies z = 2e^{i\frac{\pi}{2}}
For k=2k = 2, θ=π10+8π10=9π10  ⟹  z=2ei9π10\theta = \frac{\pi}{10} + \frac{8\pi}{10} = \frac{9\pi}{10} \implies z = 2e^{i\frac{9\pi}{10}}
For k=−1k = -1, θ=π10−4π10=−3π10  ⟹  z=2e−i3π10\theta = \frac{\pi}{10} - \frac{4\pi}{10} = -\frac{3\pi}{10} \implies z = 2e^{-i\frac{3\pi}{10}}
For k=−2k = -2, θ=π10−8π10=−7π10  ⟹  z=2e−i7π10\theta = \frac{\pi}{10} - \frac{8\pi}{10} = -\frac{7\pi}{10} \implies z = 2e^{-i\frac{7\pi}{10}}

The 5 roots are 2e−i7π10,2e−i3π10,2eiπ10,2eiπ2,2ei9π102e^{-i\frac{7\pi}{10}}, 2e^{-i\frac{3\pi}{10}}, 2e^{i\frac{\pi}{10}}, 2e^{i\frac{\pi}{2}}, 2e^{i\frac{9\pi}{10}}.

4. Sketch the roots on an Argand diagram, identifying points $P_1$ to $P_5$.

The roots represent 5 equally spaced points on a circle of radius 2 centered at the origin.
Ordering the roots by smallest argument to largest:
P1P_1 represents 2e−i7π102e^{-i\frac{7\pi}{10}}
P2P_2 represents 2e−i3π102e^{-i\frac{3\pi}{10}}
P3P_3 represents 2eiπ102e^{i\frac{\pi}{10}}
P4P_4 represents 2eiπ22e^{i\frac{\pi}{2}} (on the positive imaginary axis)
P5P_5 represents 2ei9π102e^{i\frac{9\pi}{10}}

(Self-drawn Argand diagram sketch showing a regular pentagon inscribed in a circle of radius 2, with vertices labelled P1P_1 to P5P_5 in an anticlockwise sequence.)

5. Calculate the area of the polygon $P_1 P_2 P_3 P_4 P_5$.

The polygon is a regular pentagon, which can be divided into 5 identical isosceles triangles extending from the origin.
For each triangle, the two equal sides have length r=2r = 2, and the included angle is 2π5\frac{2\pi}{5}.
Area of pentagon=5×(12r2sin⁡(2π5))\text{Area of pentagon} = 5 \times \left( \frac{1}{2} r^2 \sin\left(\frac{2\pi}{5}\right) \right)
Area=5×12(2)2sin⁡(2π5)\text{Area} = 5 \times \frac{1}{2} (2)^2 \sin\left(\frac{2\pi}{5}\right)
Area=10sin⁡(2π5)≈9.51 units2\text{Area} = 10\sin\left(\frac{2\pi}{5}\right) \approx 9.51 \text{ units}^2

Answer

Roots: 2e−i7π102e^{-i\frac{7\pi}{10}}, 2e−i3π102e^{-i\frac{3\pi}{10}}, 2eiπ102e^{i\frac{\pi}{10}}, 2eiπ22e^{i\frac{\pi}{2}}, 2ei9π102e^{i\frac{9\pi}{10}}.
Area =10sin⁡(2π5)= 10\sin\left(\frac{2\pi}{5}\right) (or ≈9.51\approx 9.51)

8(b)

1. Identify the Cartesian equations for the two individual locus conditions.

Let z=x+iyz = x + iy.
The condition ∣z−(2+2i)∣≤1|z - (2+2i)| \le 1 represents the region inside and on the circle with centre (2,2)(2, 2) and radius 11. Its Cartesian inequality is:
(x−2)2+(y−2)2≤1(x-2)^2 + (y-2)^2 \le 1

The condition arg⁡(z−1)=arg⁡(1+3i)\arg(z-1) = \arg(1+3i) represents a half-line starting from (1,0)(1,0) with gradient 33. Its Cartesian equation is:
y=3(x−1)for x>1y = 3(x-1) \quad \text{for } x > 1

2. Find the intersection of the two loci to determine the exact endpoints of the locus of $P$.

Substitute the line equation y=3x−3y = 3x - 3 into the circle's inequality:
(x−2)2+(3x−3−2)2≤1(x-2)^2 + (3x-3 - 2)^2 \le 1
(x−2)2+(3x−5)2≤1(x-2)^2 + (3x-5)^2 \le 1
x2−4x+4+9x2−30x+25≤1x^2 - 4x + 4 + 9x^2 - 30x + 25 \le 1
10x2−34x+28≤010x^2 - 34x + 28 \le 0
5x2−17x+14≤05x^2 - 17x + 14 \le 0
(5x−7)(x−2)≤0(5x-7)(x-2) \le 0
This gives 1.4≤x≤21.4 \le x \le 2.

When x=1.4x = 1.4, y=3(1.4)−3=1.2y = 3(1.4) - 3 = 1.2.
When x=2x = 2, y=3(2)−3=3y = 3(2) - 3 = 3.
The locus of PP is the solid line segment connecting the points 1.4+1.2i1.4 + 1.2i and 2+3i2 + 3i.

3. Describe the sketch of the locus of $P$ on the Argand diagram.

On an Argand diagram, indicate the circle centre (2,2)(2,2) radius 11 and the half-line from (1,0)(1,0) with dashed lines to show the boundaries.
The locus of PP is sketched by drawing a solid straight line segment from the coordinate (1.4,1.2)(1.4, 1.2) to the coordinate (2,3)(2, 3).

4. Determine the argument of $z-3-2i$ at the endpoints of the line segment locus.

Let w=z−(3+2i)w = z - (3+2i). We need the range of arg⁡(w)\arg(w) for zz on the line segment.
At the lower endpoint z1=1.4+1.2iz_1 = 1.4 + 1.2i:
w1=(1.4+1.2i)−(3+2i)=−1.6−0.8iw_1 = (1.4+1.2i) - (3+2i) = -1.6 - 0.8i
arg⁡(w1)=−π+tan⁡−1(0.81.6)=−π+tan⁡−1(0.5)≈−2.68 rad\arg(w_1) = -\pi + \tan^{-1}\left(\frac{0.8}{1.6}\right) = -\pi + \tan^{-1}(0.5) \approx -2.68 \text{ rad}

At the upper endpoint z2=2+3iz_2 = 2 + 3i:
w2=(2+3i)−(3+2i)=−1+iw_2 = (2+3i) - (3+2i) = -1 + i
arg⁡(w2)=π−tan⁡−1(11)=3π4≈2.36 rad\arg(w_2) = \pi - \tan^{-1}\left(\frac{1}{1}\right) = \frac{3\pi}{4} \approx 2.36 \text{ rad}

5. Analyse the continuous variation of the argument to determine the final range.

The segment in the ww-plane connects w1=−1.6−0.8iw_1 = -1.6 - 0.8i to w2=−1+iw_2 = -1 + i. Because the yy-coordinates transition from negative (−0.8-0.8) to positive (11) whilst the xx-coordinates remain entirely negative, the segment crosses the negative real axis (where y=0y = 0).
At the point of intersection with the negative real axis, the principal argument is π\pi.
Moving from z1z_1 to z2z_2, the argument takes all values in (−π,−π+tan⁡−1(0.5)](-\pi, -\pi + \tan^{-1}(0.5)] below the real axis, jumps to π\pi, and takes all values in [3π4,π][\frac{3\pi}{4}, \pi] above the real axis.

Therefore, the range of arg⁡(z−3−2i)\arg(z-3-2i) is:
[3π4,π]∪(−π,−π+tan⁡−1(0.5)]\left[ \frac{3\pi}{4}, \pi \right] \cup \left( -\pi, -\pi + \tan^{-1}(0.5) \right]

Answer

The locus of PP is the line segment from 1.4+1.2i1.4+1.2i to 2+3i2+3i.
Range of arg⁡(z−3−2i)\arg(z-3-2i): [3π4,π]∪(−π,−π+tan⁡−1(12)]\left[ \frac{3\pi}{4}, \pi \right] \cup \left( -\pi, -\pi + \tan^{-1}\left(\frac{1}{2}\right) \right] (or [3π4,π]∪(−π,−2.68]\left[ \frac{3\pi}{4}, \pi \right] \cup \left( -\pi, -2.68 \right] to 3 s.f.)

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