- #1
ferry2
- 15
- 0
Can you tell me is my solution true of the next problem.
Find center [tex]w_0[/tex] and radius [tex]R[/tex] of the circle [tex]k[/tex], in which the transformation [tex]w=\frac{z+2}{z-2}[/tex]
converts the line [tex]l:\text{Im} z+\text{Re} z=0[/tex].
Solution:
[tex]2 \to\infty[/tex]
[tex]-2i=(2)^*\to w_0[/tex]
[tex]w_0=w(-2i)=\frac{-2i+2}{-2i-2}=\frac{1-i}{-1-i}*\frac{-1+i}{-1+i}=i[/tex] - center of [tex]k[/tex]
[tex]0\to \frac{0+2}{0-2}=-1\in k[/tex]
[tex]R=|-1-i|=\sqrt{2}[/tex]
And can you help me with these problems:
1. Find the image of the domain [tex]\left{\begin{array}{ll}\text{Re}>0 \\ \text{Im} >0 \end{array}\right[/tex], cut along the arc [tex]\left{\begin{array}{ll} |z|=1 \\ 0 \le \arg z \le \frac{\pi}{4} \end{array}\right[/tex], by transformation [tex]w=\frac{1}{z^2}[/tex]
2. The domain [tex]\left{\begin{array}{ll} |z-1|<1 \\ |z-\frac{1}{3}|>\frac{1}{3} \end{array}\right[/tex], cut along the segment [tex][1;2][/tex], to display conformal in the stripe [tex]0<\text{Im} w<1[/tex].
Thanks in advansed :) .
Find center [tex]w_0[/tex] and radius [tex]R[/tex] of the circle [tex]k[/tex], in which the transformation [tex]w=\frac{z+2}{z-2}[/tex]
converts the line [tex]l:\text{Im} z+\text{Re} z=0[/tex].
Solution:
[tex]2 \to\infty[/tex]
[tex]-2i=(2)^*\to w_0[/tex]
[tex]w_0=w(-2i)=\frac{-2i+2}{-2i-2}=\frac{1-i}{-1-i}*\frac{-1+i}{-1+i}=i[/tex] - center of [tex]k[/tex]
[tex]0\to \frac{0+2}{0-2}=-1\in k[/tex]
[tex]R=|-1-i|=\sqrt{2}[/tex]
And can you help me with these problems:
1. Find the image of the domain [tex]\left{\begin{array}{ll}\text{Re}>0 \\ \text{Im} >0 \end{array}\right[/tex], cut along the arc [tex]\left{\begin{array}{ll} |z|=1 \\ 0 \le \arg z \le \frac{\pi}{4} \end{array}\right[/tex], by transformation [tex]w=\frac{1}{z^2}[/tex]
2. The domain [tex]\left{\begin{array}{ll} |z-1|<1 \\ |z-\frac{1}{3}|>\frac{1}{3} \end{array}\right[/tex], cut along the segment [tex][1;2][/tex], to display conformal in the stripe [tex]0<\text{Im} w<1[/tex].
Thanks in advansed :) .