image of a circle

**PeaceSoul** · March 13th 2011, 08:20 AM

can anybody please help me solve this one ? this is the last question i have ! thank you !!!

**Opalg** · March 13th 2011, 12:27 PM

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 $\dfrac{x^2}{\bigl(\frac12(c+c^{-1})\bigr)^2} + \dfrac{y^2}{\bigl(\frac12(c-c^{-1})\bigr)^2} = 1.$

**PeaceSoul** · March 13th 2011, 12:34 PM

Thank you a WHOLE lot

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