Find the image of the parabola y\,=\,x^2 under the mapping w\,= \,-2z\,+\,2i any help? Thanks
G Guest Guest May 2, 2006 #1 Find the image of the parabola \(\displaystyle y\,=\,x^2\) under the mapping \(\displaystyle w\,= \,-2z\,+\,2i\) any help? Thanks
Find the image of the parabola \(\displaystyle y\,=\,x^2\) under the mapping \(\displaystyle w\,= \,-2z\,+\,2i\) any help? Thanks
G Guest Guest Jun 9, 2006 #2 BlueFalcon said: Find the image of the parabola \(\displaystyle y\,=\,x^2\) under the mapping \(\displaystyle w\,= \,-2z\,+\,2i\) any help? Thanks Click to expand... Let \(\displaystyle z=x+iy\), then: \(\displaystyle w=-2(x+iy)+2i\) so: \(\displaystyle w=(-2x)+i(-2y+2)\) So if \(\displaystyle w=x'+iy'\), we have: \(\displaystyle x'=-2x\\ y'=-2y+2=-2x^2+2=-x'^2/2+2\) Hence the image of \(\displaystyle y\,=\,x^2\) under the given transformation is \(\displaystyle y=-x^2/2+2\) RonL
BlueFalcon said: Find the image of the parabola \(\displaystyle y\,=\,x^2\) under the mapping \(\displaystyle w\,= \,-2z\,+\,2i\) any help? Thanks Click to expand... Let \(\displaystyle z=x+iy\), then: \(\displaystyle w=-2(x+iy)+2i\) so: \(\displaystyle w=(-2x)+i(-2y+2)\) So if \(\displaystyle w=x'+iy'\), we have: \(\displaystyle x'=-2x\\ y'=-2y+2=-2x^2+2=-x'^2/2+2\) Hence the image of \(\displaystyle y\,=\,x^2\) under the given transformation is \(\displaystyle y=-x^2/2+2\) RonL
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Jun 9, 2006 #3 Hello, CaptainBlack! Welcome aboard! So you're a "New Member" . . . LOL! [At your site, I was labelled a "Newbie" for a while.]
Hello, CaptainBlack! Welcome aboard! So you're a "New Member" . . . LOL! [At your site, I was labelled a "Newbie" for a while.]
