The complex number z 1 z_1 z 1
z 1 = ( cos 5 π 12 + i sin 5 π 12 ) 4 ( cos π 3 − i sin π 3 ) 3 \begin{align*}
z_1 =\,& \frac{\left(\cos \frac{5\pi}{12} + \mathrm{i}\sin \frac{5\pi}{12}\right)^4}{\left(\cos \frac{\pi}{3} - \mathrm{i}\sin \frac{\pi}{3}\right)^3}\\[2mm]
\end{align*} z 1 = ( cos 3 π − i sin 3 π ) 3 ( cos 12 5 π + i sin 12 5 π ) 4 (a) Without using your calculator show that
z 1 = cos 2 π 3 + i sin 2 π 3 \begin{align*}
z_1 =\,& \cos \frac{2\pi}{3} + \mathrm{i}\sin \frac{2\pi}{3}\\[2mm]
\end{align*} z 1 = cos 3 2 π + i sin 3 2 π (4)
(b) Shade, on a single Argand diagram, the region R R R
∣ z − z 1 ∣ ⩽ 1 and 0 ⩽ arg ( z − z 1 ) ⩽ 3 π 4 \begin{align*}
|z - z_1| \leqslant 1 \qquad \text{and} \qquad 0 \leqslant \arg(z - z_1) \leqslant \frac{3\pi}{4}\\[2mm]
\end{align*} ∣ z − z 1 ∣ ⩽ 1 and 0 ⩽ arg ( z − z 1 ) ⩽ 4 3 π (4)
Given that the complex number z z z R R R
(c) determine the smallest possible positive value of arg z \arg z arg z
(2)
解答
(a)
For the numerator, using De Moivre’s theorem:
( cos 5 π 12 + i sin 5 π 12 ) 4 = cos ( 4 × 5 π 12 ) + i sin ( 4 × 5 π 12 ) = cos 5 π 3 + i sin 5 π 3 \begin{align*}
\left(\cos\frac{5\pi}{12} + \mathrm{i}\sin\frac{5\pi}{12}\right)^4
= &\,\cos\left(4 \times \frac{5\pi}{12}\right) + \mathrm{i}\sin\left(4 \times \frac{5\pi}{12}\right)\\[4mm]
= &\,\cos\frac{5\pi}{3} + \mathrm{i}\sin\frac{5\pi}{3}
\end{align*} ( cos 12 5 π + i sin 12 5 π ) 4 = = cos ( 4 × 12 5 π ) + i sin ( 4 × 12 5 π ) cos 3 5 π + i sin 3 5 π For the denominator, we can rewrite it first as:
cos π 3 − i sin π 3 = cos ( − π 3 ) + i sin ( − π 3 ) \begin{align*}
\cos\frac{\pi}{3} - \mathrm{i}\sin\frac{\pi}{3}
= &\,\cos\left(-\frac{\pi}{3}\right) + \mathrm{i}\sin\left(-\frac{\pi}{3}\right)
\end{align*} cos 3 π − i sin 3 π = cos ( − 3 π ) + i sin ( − 3 π ) Applying De Moivre’s theorem:
( cos ( − π 3 ) + i sin ( − π 3 ) ) 3 = cos ( − π ) + i sin ( − π ) \begin{align*}
\left(\cos\left(-\frac{\pi}{3}\right) + \mathrm{i}\sin\left(-\frac{\pi}{3}\right)\right)^3
= &\,\cos(-\pi) + \mathrm{i}\sin(-\pi)
\end{align*} ( cos ( − 3 π ) + i sin ( − 3 π ) ) 3 = cos ( − π ) + i sin ( − π ) Now, dividing the numerator by the denominator:
z 1 = cos 5 π 3 + i sin 5 π 3 cos ( − π ) + i sin ( − π ) = cos ( 5 π 3 − ( − π ) ) + i sin ( 5 π 3 − ( − π ) ) = cos ( 8 π 3 ) + i sin ( 8 π 3 ) \begin{align*}
z_1
= &\,\frac{\cos\frac{5\pi}{3} + \mathrm{i}\sin\frac{5\pi}{3}}{\cos(-\pi) + \mathrm{i}\sin(-\pi)}\\[4mm]
= &\,\cos\left(\frac{5\pi}{3} - (-\pi)\right) + \mathrm{i}\sin\left(\frac{5\pi}{3} - (-\pi)\right)\\[4mm]
= &\,\cos\left(\frac{8\pi}{3}\right) + \mathrm{i}\sin\left(\frac{8\pi}{3}\right)
\end{align*} z 1 = = = cos ( − π ) + i sin ( − π ) cos 3 5 π + i sin 3 5 π cos ( 3 5 π − ( − π ) ) + i sin ( 3 5 π − ( − π ) ) cos ( 3 8 π ) + i sin ( 3 8 π ) Subtracting 2 π 2\pi 2 π
8 π 3 − 2 π = 2 π 3 \begin{align*}
\frac{8\pi}{3} - 2\pi
= &\,\frac{2\pi}{3}
\end{align*} 3 8 π − 2 π = 3 2 π Thus, z 1 = cos 2 π 3 + i sin 2 π 3 z_1 = \cos\frac{2\pi}{3} + \mathrm{i}\sin\frac{2\pi}{3} z 1 = cos 3 2 π + i sin 3 2 π
(b)
Sketch instruction: Draw an Argand diagram. Plot the point z 1 = cos 2 π 3 + i sin 2 π 3 z_1 = \cos\frac{2\pi}{3} + \mathrm{i}\sin\frac{2\pi}{3} z 1 = cos 3 2 π + i sin 3 2 π 1 1 1 z 1 z_1 z 1 z 1 z_1 z 1 arg ( z − z 1 ) = 0 \arg(z-z_1) = 0 arg ( z − z 1 ) = 0 3 π 4 \frac{3\pi}{4} 4 3 π 0 0 0 3 π 4 \frac{3\pi}{4} 4 3 π
(c)
The complex number z z z R R R z 1 z_1 z 1
z 1 = − 1 2 + i 3 2 \begin{align*}
z_1
= &\,-\frac{1}{2} + \mathrm{i}\frac{\sqrt{3}}{2}
\end{align*} z 1 = − 2 1 + i 2 3 The minimum value of arg z \arg z arg z R R R 0 0 0
z = z 1 + 1 = ( − 1 2 + i 3 2 ) + 1 = 1 2 + i 3 2 \begin{align*}
z
= &\,z_1 + 1\\[4mm]
= &\,\left(-\frac{1}{2} + \mathrm{i}\frac{\sqrt{3}}{2}\right) + 1\\[4mm]
= &\,\frac{1}{2} + \mathrm{i}\frac{\sqrt{3}}{2}
\end{align*} z = = = z 1 + 1 ( − 2 1 + i 2 3 ) + 1 2 1 + i 2 3 The argument of this point is:
arg z = arctan ( 3 / 2 1 / 2 ) = arctan ( 3 ) = π 3 \begin{align*}
\arg z
= &\,\arctan\left(\frac{\sqrt{3}/2}{1/2}\right)\\[4mm]
= &\,\arctan(\sqrt{3})\\[4mm]
= &\,\frac{\pi}{3}
\end{align*} arg z = = = arctan ( 1/2 3 /2 ) arctan ( 3 ) 3 π 