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    Consider $$w = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$$

    Question
    HLPaper 1

    Consider w=2(cos⁡π3+isin⁡π3)w = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)w=2(cos3π​+isin3π​)

    These four points form the vertices of a quadrilateral, Q.

    1.

    Express w2w^2w2 and w3w^3w3 in modulus-argument form.

    [3]
    Verified
    Solution

    w2=4cis(2π3);  w3=8cis(π)w^2 = 4\text{cis}\left( {\frac{{2\pi }}{3}} \right);\,\,w^3 = 8{\text{cis}}\left( \pi \right)w2=4cis(32π​);w3=8cis(π) M1 can be awarded for either both correct moduli or both correct arguments. A1A1

    Accept Euler form. Allow multiplication of correct Cartesian form for M1, final answers must be in modulus-argument form.

    2.

    Sketch on an Argand diagram the points represented by w0w^0w0, w1w^1w1, w2w^2w2 and w3w^3w3.

    [2]
    Verified
    Solution

    Markscheme

    A1 for correctly sketching w0=1w^0 = 1w0=1 on the positive real axis

    A1 for correctly sketching the following points:

    • w1=2cos⁡(π3)+2isin⁡(π3)w^1 = 2\cos\left(\frac{\pi}{3}\right) + 2i\sin\left(\frac{\pi}{3}\right)w1=2cos(3π​)+2isin(3π​)
    • w2=4cos⁡(2π3)+4isin⁡(2π3)w^2 = 4\cos\left(\frac{2\pi}{3}\right) + 4i\sin\left(\frac{2\pi}{3}\right)w2=4cos(32π​)+4isin(32π​)
    • w3=−8w^3 = -8w3=−8
    Re∙Re↶∙∙↷Im∙Im\begin{array}{ccc} \text{Re} & \bullet & \text{Re}\\ & \curvearrowleft & \\ & \bullet & \bullet\\ & \curvearrowright & \\ \text{Im} & \bullet & \text{Im} \end{array}ReIm​∙↶∙↷∙​Re∙Im​ 2
    3.

    Show that the area of the quadrilateral Q is 2132\frac{21\sqrt{3}}{2}2213​​.

    [3]
    Verified
    Solution

    Method #1

    Area=12absin⁡C\text{Area} = \frac{1}{2}ab\sin CArea=21​absinC

    M1 for using the area formula

    12×1×2×sin⁡π3+12×2×4×sin⁡π3+12×4×8×sin⁡π3\frac{1}{2} \times 1 \times 2 \times \sin\frac{\pi}{3} + \frac{1}{2} \times 2 \times 4 \times \sin\frac{\pi}{3} + \frac{1}{2} \times 4 \times 8 \times \sin\frac{\pi}{3}21​×1×2×sin3π​+21​×2×4×sin3π​+21​×4×8×sin3π​

    A1 for C=π3C = \frac{\pi}{3}C=3π​ A1 for correct moduli

    =2132= \frac{21\sqrt{3}}{2}=2213​​

    AG

    Other methods of splitting the area may receive full marks.

    3

    4.

    Let z=2(cos⁡πn+isin⁡πn)z = 2(\cos\frac{\pi}{n} + i\sin\frac{\pi}{n})z=2(cosnπ​+isinnπ​), n∈Z+n \in \mathbb{Z}^+n∈Z+. The points represented on an Argand diagram by z0z^0z0, z1z^1z1, z2z^2z2, …\ldots…, znz^nzn form the vertices of a polygon PnP_nPn​.

    Show that the area of the polygon PnP_nPn​ can be expressed in the form a(bn−1)sin⁡πna(b^n - 1)\sin\frac{\pi}{n}a(bn−1)sinnπ​, where a,b∈Ra, b \in \mathbb{R}a,b∈R.

    [6]
    Verified
    Solution

    Method #1

    12×20×21×sinπn+12×21×22×sinπn+12×22×23×sinπn+ … +12×2n−1×2n×sinπn\frac{1}{2} \times {2^0} \times {2^1} \times {\text{sin}}\frac{\pi}{n} + \frac{1}{2} \times {2^1} \times {2^2} \times {\text{sin}}\frac{\pi}{n} + \frac{1}{2} \times {2^2} \times {2^3} \times {\text{sin}}\frac{\pi}{n} + \, \ldots \, + \frac{1}{2} \times {2^{n - 1}} \times {2^n} \times {\text{sin}}\frac{\pi}{n}21​×20×21×sinnπ​+21​×21×22×sinnπ​+21​×22×23×sinnπ​+…+21​×2n−1×2n×sinnπ​

    M1A1

    Award M1 for powers of 2, A1 for any correct expression including both the first and last term.

    =sinπn×(20+22+24+ … +2n−2)= {\text{sin}}\frac{\pi}{n} \times \left( {{2^0} + {2^2} + {2^4} + \, \ldots \, + {2^{n - 2}}} \right)=sinnπ​×(20+22+24+…+2n−2)

    Identifying a geometric series with common ratio 22 (= 4) M1A1

    =1−22n1−4×sinπn= \frac{{1 - {2^{2n}}}}{{1 - 4}} \times {\text{sin}}\frac{\pi}{n}=1−41−22n​×sinnπ​

    M1

    Award M1 for use of formula for sum of geometric series.

    =13(4n−1)sinπn= \frac{1}{3}\left( {{4^n} - 1} \right){\text{sin}}\frac{\pi}{n}=31​(4n−1)sinnπ​

    A1

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