z be a variable complex number such that |z|=2.Show that the point z+1/z lies on an ellipse of eccentricity 4/5 in the complex plane.

Question

I started with the formula $|z+1/z|^2=|z|^2+|1/z|^2+2Re(z×{1 \over \bar z})$. Then I calculated but I didn't get the answer.I got $e=\sqrt{14 \over 23}$.

Answer 1

$z=x+yi$ then $$ \left|z+\frac{1}{z}\right|^2 = \left| z+\frac{z^*}{zz^*}\right|^2 = \left| z+\frac{z^*}{4}\right|^2 = \left| (5/4)x+(3/4)yi\right|^2 = (5/4)^2x^2+(3/4)^2y^2 $$ $$ e=\sqrt{1-\left(\frac{3/4}{5/4}\right)^2} = \frac{4}{5} $$

Answer 2

We can write $z=2(\cos\theta+i\sin\theta)$, as $|z|=2$.

Hence, $z+{1 \over z}=(2\cos\theta+{1 \over 2}\cos\theta)+i(2\sin\theta-{1 \over 2}\sin\theta)$.

If we look at the complex plane, then this represents the point $((2+{1 \over 2})\cos\theta,(2-{1 \over 2})\sin\theta)\equiv({5 \over 2}\cos\theta,{3 \over 2}\sin\theta)$. If we denote this by $(x,y)$, then it's easy to see that ${4x^2 \over 25}+{4y^2 \over 9}=1$.

That's the equation of an ellipse with eccentricity $e=\sqrt{1-{9 \over 25}}={4 \over 5}$.

Answer 3

Making $z=\rho e^{i\phi}$ we have

$$ |z|=2 \Rightarrow \rho = 2\to w = z+\frac 1z = 2e^{i\phi}+\frac 12 e^{-i\phi} $$

or

$$ w = \frac 12\left(4\cos\phi+4i8\sin\phi+\cos\phi-i\sin\phi\right) = \frac 12\left(5\cos\phi + 3i\sin\phi\right) $$

now calling $2w = X + i Y\; $ we have

$$ X = 5\cos\phi\\ Y = 3\sin\phi $$

or

$$ \left(\frac X5\right)^2+\left(\frac Y3\right)^2 = 1 $$

etc.