The complex numbers \(w\) and \(z\) are given by \(w=2 e^{\small{\dfrac{i \pi}{6}}}\) and \(z=3 e^{\small{\dfrac{i \pi}{6}}}\). Find the modulus and argument of \(w z\). (2 marks)
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Complex Numbers, EXT2 N1 EQ-Bank 9
Calculate the value of \(\dfrac{e^{\small{\dfrac{i \pi}{3}}}-e^{-\small{\dfrac{i \pi}{3}}}}{2 i}\). (2 marks)
Show Worked Solution
| \(\dfrac{e^{\small{\dfrac{i \pi}{3}}}-e^{-\small{\dfrac{i \pi}{3}}}}{2 i}\) | \(=\dfrac{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}-\left(\cos \left(-\frac{\pi}{3}\right)+i \sin \left(-\frac{\pi}{3}\right)\right)}{2 i}\) |
| \(=\dfrac{\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}-\left(\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}\right)}{2 i}\) | |
| \(=\dfrac{2 i \sin \frac{\pi}{3}}{2 i}\) | |
| \(=\sin \frac{\pi}{3}\) | |
| \(=\dfrac{\sqrt{3}}{2}\) |
Complex Numbers, EXT2 N1 EQ-Bank 6
Given \(z=\dfrac{-1-i \sqrt{3}}{1+i}\), calculate \(z^2\) in exponential form. (3 marks)
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Show Worked Solution
| \(z\) | \(=\dfrac{-1-i \sqrt{3}}{1+i} \times \dfrac{1-i}{1-i}\) |
| \(=\dfrac{(-1-i \sqrt{3})(1-i)}{1-i^2}\) | |
| \(=\dfrac{-1+i-i \sqrt{3}+i^2 \sqrt{3}}{2}\) | |
| \(=\dfrac{-1-\sqrt{3}}{2}+i\left(\dfrac{1-\sqrt{3}}{2}\right)\) |
| \(\abs{z}\) | \(=\sqrt{\left(\dfrac{-1-\sqrt{3}}{2}\right)^2+\left(\dfrac{1-\sqrt{3}}{2}\right)^2}\) |
| \(=\sqrt{\dfrac{1+2 \sqrt{3}+3+1-2 \sqrt{3}+3}{4}}\) | |
| \(=\sqrt{2}\) |
\(\text{Find}\ \ \arg (z):\)
| \(\tan \theta\) | \(=\dfrac{\frac{1-\sqrt{3}}{2}}{\frac{-1-\sqrt{3}}{2}}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\) |
| \(\theta\) | \(=15^{\circ}=\dfrac{\pi}{12}\) |
\(\therefore \arg (z)=-\dfrac{11 \pi}{12}\)
| \(z\) | \(=\sqrt{2} \operatorname{cis}\left(-\dfrac{11 \pi}{12}\right)\) |
| \(z^2\) | \(=(\sqrt{2})^2 \operatorname{cis}\left(-\dfrac{11 \pi}{12} \times 2+2 \pi\right)\) |
| \(=2 \operatorname{cis}\left(\dfrac{\pi}{6}\right)\) | |
| \(=2 e^{\small{\dfrac{i \pi}{6}}}\) |