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 y = x2\displaystyle y\,=\,x^2y=x2 under the mapping w = −2z + 2i\displaystyle w\,= \,-2z\,+\,2iw=−2z+2i any help? Thanks
Find the image of the parabola y = x2\displaystyle y\,=\,x^2y=x2 under the mapping w = −2z + 2i\displaystyle w\,= \,-2z\,+\,2iw=−2z+2i any help? Thanks
G Guest Guest Jun 9, 2006 #2 BlueFalcon said: Find the image of the parabola y = x2\displaystyle y\,=\,x^2y=x2 under the mapping w = −2z + 2i\displaystyle w\,= \,-2z\,+\,2iw=−2z+2i any help? Thanks Click to expand... Let z=x+iy\displaystyle z=x+iyz=x+iy, then: w=−2(x+iy)+2i\displaystyle w=-2(x+iy)+2iw=−2(x+iy)+2i so: w=(−2x)+i(−2y+2)\displaystyle w=(-2x)+i(-2y+2)w=(−2x)+i(−2y+2) So if w=x′+iy′\displaystyle w=x'+iy'w=x′+iy′, we have: \(\displaystyle x'=-2x\\ y'=-2y+2=-2x^2+2=-x'^2/2+2\) Hence the image of y = x2\displaystyle y\,=\,x^2y=x2 under the given transformation is y=−x2/2+2\displaystyle y=-x^2/2+2y=−x2/2+2 RonL
BlueFalcon said: Find the image of the parabola y = x2\displaystyle y\,=\,x^2y=x2 under the mapping w = −2z + 2i\displaystyle w\,= \,-2z\,+\,2iw=−2z+2i any help? Thanks Click to expand... Let z=x+iy\displaystyle z=x+iyz=x+iy, then: w=−2(x+iy)+2i\displaystyle w=-2(x+iy)+2iw=−2(x+iy)+2i so: w=(−2x)+i(−2y+2)\displaystyle w=(-2x)+i(-2y+2)w=(−2x)+i(−2y+2) So if w=x′+iy′\displaystyle w=x'+iy'w=x′+iy′, we have: \(\displaystyle x'=-2x\\ y'=-2y+2=-2x^2+2=-x'^2/2+2\) Hence the image of y = x2\displaystyle y\,=\,x^2y=x2 under the given transformation is y=−x2/2+2\displaystyle y=-x^2/2+2y=−x2/2+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.]
