<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Show that $1-{{\mathrm{e}}^{\mathrm{i}n\theta }}={{\mathrm{e}}^{\frac{1}{2}\mathrm{i}n\theta }}\left( -2\mathrm{i}\sin \frac{n\theta }{2} \right)$, and $1+{{\mathrm{e}}^{\mathrm{i}n\theta }}={{\mathrm{e}}^{\frac{1}{2}\mathrm{i}n\theta }}\left( 2\cos \frac{n\theta }{2} \right)$ where $n\in \mathbb{R}$.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">It is given that $z={{\mathrm{e}}^{\mathrm{i}\theta }}$, where $0\le \theta \le \frac{\pi }{2}$, using (i), show that</span></div><div class="list-item-paragraph" style="text-align: center">$\left| 1-z+{{z}^{2}}-{{z}^{3}} \right|=4\sin \frac{\theta }{2}\cos \theta$.</div><p style="text-align: right;">[4]</p></li></ol>

Show that $1-{{\mathrm{e}}^{\mathrm{i}n\theta }}={{\mathrm{e}}^{\frac{1}{2}\mathrm{i}n\theta }}\left( -2\mathrm{i}\sin \frac{n\theta }{2} \right)$, and $1+{{\mathrm{e}}^{\mathrm{i}n\theta }}={{\mathrm{e}}^{\frac{1}{2}\mathrm{i}n\theta }}\left( 2\cos \frac{n\theta }{2} \right)$ where $n\in \mathbb{R}$.

It is given that $z={{\mathrm{e}}^{\mathrm{i}\theta }}$, where $0\le \theta \le \frac{\pi }{2}$, using (i), show that

$\left| 1-z+{{z}^{2}}-{{z}^{3}} \right|=4\sin \frac{\theta }{2}\cos \theta$.

Complex Numbers - General

Complex Numbers

International Baccalaureate Higher Level Mathematics

2018 MJC P1 Q3

2018 MJC P1 Q3

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