A transformation T T T z z z z = x + i y z = x + iy z = x + i y w w w w = u + i v w = u + iv w = u + i v
w = ( 3 − i ) ( z − 2 ) z + 2 z ≠ − 2 \begin{align*}
w = \frac{(\sqrt{3} - \mathrm{i})(z - 2)}{z + 2} \qquad z \neq -2
\end{align*} w = z + 2 ( 3 − i ) ( z − 2 ) z = − 2 (a) Show that the real axis in the z z z T T T
v = − 1 3 u \begin{align*}
v = -\frac{1}{\sqrt{3}}u
\end{align*} v = − 3 1 u in the w w w
(3)
(b) Show that the circle in the z z z ∣ z ∣ = 2 |z| = 2 ∣ z ∣ = 2 T T T w w w
(5)
The region R R R z z z
{ z ∈ C : ∣ z ∣ < 2 } ∩ { z ∈ C : I m z > 0 } \begin{align*}
\{z \in \mathbb{C} : |z| < 2\} \cap \{z \in \mathbb{C} : \mathrm{Im}\,z > 0\}
\end{align*} { z ∈ C : ∣ z ∣ < 2 } ∩ { z ∈ C : Im z > 0 } (c) Determine the image of R R R T T T
{ w ∈ C : α < arg w < β } \begin{align*}
\{w \in \mathbb{C} : \alpha < \arg w < \beta\}
\end{align*} { w ∈ C : α < arg w < β } where α \alpha α β \beta β π \pi π
(5)
解答
(a)
The real axis in the z z z y = 0 y = 0 y = 0 z = x z = x z = x
w = ( 3 − i ) ( x − 2 ) x + 2 u + i v = 3 ( x − 2 ) x + 2 − i x − 2 x + 2 \begin{align*}
w
= &\,\frac{(\sqrt{3} - \mathrm{i})(x - 2)}{x + 2}\\[4mm]
u + \mathrm{i}v
= &\,\frac{\sqrt{3}(x - 2)}{x + 2} - \mathrm{i}\frac{x - 2}{x + 2}
\end{align*} w = u + i v = x + 2 ( 3 − i ) ( x − 2 ) x + 2 3 ( x − 2 ) − i x + 2 x − 2 Equating the real and imaginary parts:
u = 3 ( x − 2 ) x + 2 v = − x − 2 x + 2 \begin{align*}
u
= &\,\frac{\sqrt{3}(x - 2)}{x + 2}\\[4mm]
v
= &\,-\frac{x - 2}{x + 2}
\end{align*} u = v = x + 2 3 ( x − 2 ) − x + 2 x − 2 Dividing u u u v v v
u v = 3 − 1 u = − 3 v v = − 1 3 u \begin{align*}
\frac{u}{v}
= &\,\frac{\sqrt{3}}{-1}\\[4mm]
u
= &\,-\sqrt{3}v\\[4mm]
v
= &\,-\frac{1}{\sqrt{3}}u
\end{align*} v u = u = v = − 1 3 − 3 v − 3 1 u
(b)
Rearranging the given transformation to make z z z
w ( z + 2 ) = ( 3 − i ) ( z − 2 ) w z + 2 w = ( 3 − i ) z − 2 ( 3 − i ) z ( w − 3 + i ) = − 2 w − 2 ( 3 − i ) \begin{align*}
w(z + 2)
= &\,(\sqrt{3} - \mathrm{i})(z - 2)\\[4mm]
wz + 2w
= &\,(\sqrt{3} - \mathrm{i})z - 2(\sqrt{3} - \mathrm{i})\\[4mm]
z(w - \sqrt{3} + \mathrm{i})
= &\,-2w - 2(\sqrt{3} - \mathrm{i})
\end{align*} w ( z + 2 ) = w z + 2 w = z ( w − 3 + i ) = ( 3 − i ) ( z − 2 ) ( 3 − i ) z − 2 ( 3 − i ) − 2 w − 2 ( 3 − i ) Since ∣ z ∣ = 2 |z| = 2 ∣ z ∣ = 2
∣ z ∣ = ∣ − 2 w − 2 3 + 2 i w − 3 + i ∣ = 2 2 ∣ w + 3 − i ∣ ∣ w − 3 + i ∣ = 2 ∣ w + 3 − i ∣ = ∣ w − 3 + i ∣ \begin{align*}
|z|
= &\,\left| \frac{-2w - 2\sqrt{3} + 2\mathrm{i}}{w - \sqrt{3} + \mathrm{i}} \right|
= 2\\[4mm]
\frac{2|w + \sqrt{3} - \mathrm{i}|}{|w - \sqrt{3} + \mathrm{i}|}
= &\,2\\[4mm]
|w + \sqrt{3} - \mathrm{i}|
= &\,|w - \sqrt{3} + \mathrm{i}|
\end{align*} ∣ z ∣ = ∣ w − 3 + i ∣ 2∣ w + 3 − i ∣ = ∣ w + 3 − i ∣ = w − 3 + i − 2 w − 2 3 + 2 i = 2 2 ∣ w − 3 + i ∣ This represents the perpendicular bisector of the line segment joining ( − 3 , 1 ) (-\sqrt{3}, 1) ( − 3 , 1 ) ( 3 , − 1 ) (\sqrt{3}, -1) ( 3 , − 1 ) w = u + i v w = u + \mathrm{i}v w = u + i v
( u + 3 ) 2 + ( v − 1 ) 2 = ( u − 3 ) 2 + ( v + 1 ) 2 u 2 + 2 3 u + 3 + v 2 − 2 v + 1 = u 2 − 2 3 u + 3 + v 2 + 2 v + 1 4 3 u = 4 v v = 3 u \begin{align*}
(u + \sqrt{3})^2 + (v - 1)^2
= &\,(u - \sqrt{3})^2 + (v + 1)^2\\[4mm]
u^2 + 2\sqrt{3}u + 3 + v^2 - 2v + 1
= &\,u^2 - 2\sqrt{3}u + 3 + v^2 + 2v + 1\\[4mm]
4\sqrt{3}u
= &\,4v\\[4mm]
v
= &\,\sqrt{3}u
\end{align*} ( u + 3 ) 2 + ( v − 1 ) 2 = u 2 + 2 3 u + 3 + v 2 − 2 v + 1 = 4 3 u = v = ( u − 3 ) 2 + ( v + 1 ) 2 u 2 − 2 3 u + 3 + v 2 + 2 v + 1 4 v 3 u
(c)
The boundary components of region R R R v = − 1 3 u ⟹ arg w = − π 6 v = -\frac{1}{\sqrt{3}}u \implies \arg w = -\frac{\pi}{6} v = − 3 1 u ⟹ arg w = − 6 π 5 π 6 \frac{5\pi}{6} 6 5 π v = 3 u ⟹ arg w = π 3 v = \sqrt{3}u \implies \arg w = \frac{\pi}{3} v = 3 u ⟹ arg w = 3 π − 2 π 3 -\frac{2\pi}{3} − 3 2 π
To find the correct sector mapped from R R R R R R z = 0 z = 0 z = 0 R R R ∣ 0 ∣ < 2 , y = 0 |0|<2, y=0 ∣0∣ < 2 , y = 0
w = ( 3 − i ) ( 0 − 2 ) 0 + 2 = − 3 + i \begin{align*}
w
= &\,\frac{(\sqrt{3} - \mathrm{i})(0 - 2)}{0 + 2}\\[4mm]
= &\,-\sqrt{3} + \mathrm{i}
\end{align*} w = = 0 + 2 ( 3 − i ) ( 0 − 2 ) − 3 + i The argument of − 3 + i -\sqrt{3} + \mathrm{i} − 3 + i 5 π 6 \frac{5\pi}{6} 6 5 π R R R I m z > 0 \mathrm{Im} \ z > 0 Im z > 0 ∣ z ∣ < 2 |z|<2 ∣ z ∣ < 2 5 π 6 \frac{5\pi}{6} 6 5 π π 3 \frac{\pi}{3} 3 π
Therefore, the image of R R R T T T
{ w ∈ C : π 3 < arg w < 5 π 6 } \begin{align*}
\left\{w \in \mathbb{C} : \frac{\pi}{3} < \arg w < \frac{5\pi}{6}\right\}
\end{align*} { w ∈ C : 3 π < arg w < 6 5 π } 