# image of a circle

• March 13th 2011, 08:20 AM
PeaceSoul
image of a circle
Attachment 21133

can anybody please help me solve this one ? this is the last question i have ! thank you !!!
• March 13th 2011, 12:27 PM
Opalg
If you put $z = ce^{i\theta}$ (so that $z=x+iy$ lies on the circle $|z|=c$), then

$w = \frac12(z+z^{-1}) = \frac12(ce^{i\theta} + c^{-1}e^{-i\theta}) = \frac12(c+c^{-1})\cos\theta + \frac i2(c-c^{-1})\sin\theta.$

So w lies on the ellipse $\dfrac{x^2}{\bigl(\frac12(c+c^{-1})\bigr)^2} + \dfrac{y^2}{\bigl(\frac12(c-c^{-1})\bigr)^2} = 1.$
• March 13th 2011, 12:34 PM
PeaceSoul
Thank you a WHOLE lot