IB Maths AA HL 1.13 Polar and Euler forms Question Bank

Practise IB Mathematics HL 1.13 by applying polar and euler forms methods to exam-style questions.

[Maximum number: 9]

Let $z=1-\cos 2 \theta-\mathrm{i} \sin 2 \theta, z \in \mathbb{C}, 0 \leq \theta \leq \pi$ .

Find the modulus and argument of z in terms of $\theta$ . Express each answer in its simplest form.

EITHER

\begin{array}{ll} z=(1-\cos 2 \theta)-i \sin 2 \theta & \\ |z|=\sqrt{(1-\cos 2 \theta)^{2}+(\sin 2 \theta)^{2}} & \boldsymbol{M} \mathbf{1} \\ |z|=\sqrt{1-2 \cos 2 \theta+\cos ^{2} 2 \theta+\sin ^{2} 2 \theta} & \boldsymbol{A} \mathbf{1} \\ =\sqrt{2} \sqrt{(1-\cos 2 \theta)} & \boldsymbol{A} \mathbf{1} \\ =\sqrt{2\left(2 \sin ^{2} \theta\right)} & \\ =2 \sin \theta & \boldsymbol{A} \mathbf{1} \end{array}

let $\arg (z)=\alpha$ $\tan \alpha=-\frac{\sin 2 \theta}{1-\cos 2 \theta}$ $=\frac{-2 \sin \theta \cos \theta}{2 \sin ^{2} \theta}$ $=-\cot \theta$ $\arg (z)=\alpha=-\arctan \left(\tan \left(\frac{\pi}{2}-\theta\right)\right)$ $=\theta-\frac{\pi}{2}$
OR
$z=(1-\cos 2 \theta)-\mathrm{i} \sin 2 \theta$ $=2 \sin ^{2} \theta-2 i \sin \theta \cos \theta$ $=2 \sin \theta(\sin \theta-i \cos \theta)$ $=-2 i \sin \theta(\cos \theta+i \sin \theta)$ $=2 \sin \theta\left(\cos \left(\theta-\frac{\pi}{2}\right)+i \sin \left(\theta-\frac{\pi}{2}\right)\right)$ $|z|=2 \sin \theta$ $\arg (z)=\theta-\frac{\pi}{2}$

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AHL 1.13 (HL)—Polar and Euler forms question 1