A-Level Mathematics May/June 2024 Q9(b): Express z and ω in the form re^(iθ), where r > 0 and −π < θ ≤ π. Give the exact values…

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  For the complex number z, which from part (a) is 1 - i:

  The modulus is $r = |z| = \sqrt{1^2 + (-1)^2} = \sqrt{1+1} = \sqrt{2}$.

  The argument $\theta = \arg(z)$ is found by considering the position of (1, -1) on an Argand diagram. It is in the fourth quadrant. The basic angle is $\alpha = \tan^{-1}\left(\frac{|-1|}{|1|}\right) = \tan^{-1}(1) = \frac{\pi}{4}$. For the fourth quadrant, the argument is $\theta = -\alpha = -\frac{\pi}{4}$.

  Therefore, $z = \sqrt{2}e^{-i\pi/4}$.

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  For the complex number $\omega$, which from part (a) is $-3 + 3i\sqrt{3}$:

  The modulus is $r = |\omega| = \sqrt{(-3)^2 + (3\sqrt{3})^2} = \sqrt{9 + 27} = \sqrt{36} = 6$.

  The argument $\theta = \arg(\omega)$ is found by considering the position of $(-3, 3\sqrt{3})$ on an Argand diagram. It is in the second quadrant. The basic angle is $\alpha = \tan^{-1}\left(\frac{|3\sqrt{3}|}{|-3|}\right) = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$. For the second quadrant, the argument is $\theta = \pi - \alpha = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$.

  Therefore, $\omega = 6e^{i2\pi/3}$.

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  How the marks are awarded

  • B1 — Correctly calculating the modulus of z as $\sqrt{2}$.
  • B1 — Correctly calculating the argument of z as $-\frac{\pi}{4}$. This is a final answer mark.
  • B1 — Correctly calculating the modulus of ω as 6.
  • B1 — Correctly calculating the argument of ω as $\frac{2\pi}{3}$. This is a final answer mark.

  Common mistakes

  • Quadrant error for the argument of ω: calculating the basic angle $\frac{\pi}{3}$ and using this as the argument, instead of correctly finding the second-quadrant angle $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$.
  • Argument range error: giving the argument of z as $\frac{7\pi}{4}$ instead of $-\frac{pi}{4}$, which is outside the required range $-\pi < \theta \leq \pi$.
  • Calculator mode error: working in degrees and stating the arguments as $-45^\circ$ and $120^\circ$ without converting to exact radians as required.
  • Arithmetic error in modulus calculation: for ω, incorrectly squaring the real part, e.g., $\sqrt{-3^2 + ...} = \sqrt{-9 + ...}$, which leads to an incorrect or impossible modulus.

  Examiner tip: Always sketch a quick Argand diagram to correctly identify the quadrant for the argument, ensuring your final angle is within the specified principal range.

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