$$ \text{Find the locus of $w$, where $z$ is restricted as indicated:} \\ w = z - \frac{1}{z} \\ \text{if } |z| = 2 $$
I have tried solving this by multiplying both sides by $z$, and then using the quadtratic equation. I get $z = \frac{w \pm \sqrt{w^2+4}}{2}$. I then set $0 \leq w^2+4 $ But I still have no idea on how to solve this.
Thanks in advance.
It's an ellipse.
Write $z=2\cos\theta+2i\sin\theta$.
Then $\dfrac 1 z = \dfrac 1 2 (\cos\theta-i\sin\theta)$ so $$ z -\frac 1 z = \frac 3 2 \cos\theta + \frac 5 2 i\sin\theta = x + i y $$ where $x,y$ are real. Then we have $$ \frac{x^2}{(3/2)^2} + \frac{y^2}{(5/2)^2} = 1. $$
let $z = 2(\cos t + i \sin t)$. Then $$w= u + iv= z - \frac 1 z=2(\cos t + i \sin t ) - \frac 1 2(\cos t - i\sin t). $$ that gives you $$u = \frac 3 2\cos t, v = \frac 52 \sin t, \, \text{ which is an ellipse } \frac 49 u^2 + \frac 4 {25} v^2 = 1. $$
