Given $w + 2i = z + \frac{1}{z}$

Let $z = re^{i\theta}$ and w = u + iv

u + iv + 2i $= re^{i\theta} + \frac{1}{re^{i\theta}} \\ = r(cos\theta + isin\theta) + \frac{1}{r} e^{-i\theta} \\ = r(cos\theta + isin\theta) + \frac{1}{r}(cos\theta - isin\theta)$

u + i(v + 2) = $\bigg(r + \frac{1}{r}\bigg)cos\theta + i\bigg(r - \frac{1}{r}\bigg)sin\theta$

$u = \bigg(r + \frac{1}{r}\bigg)cos\theta$

$v = \bigg(r - \frac{1}{r}\bigg)cos\theta$

$\therefore \frac{u}{cos\theta} = r + \frac{1}{r}, \frac{v}{sin\theta} = r - \frac{1}{r}$

i.e.

$\frac{u}{r + \frac{1}{r}} = cos\theta , \frac{v}{r - \frac{1}{r}} =sin\theta$

Squaring of both the equations

$\frac{u^2}{\bigg(r + \frac{1}{r}\bigg)^2} = cos^2\theta...........(1)$

$\frac{v^2}{\bigg(r - \frac{1}{r}\bigg)^2} = sin^2\theta..............(2)$

Adding equations (1) and (2)

$\frac{u^2}{\bigg(r + \frac{1}{r}\bigg)^2} + \frac{v^2}{\bigg(r - \frac{1}{r}\bigg)^2} = 1........(3)$

given |z| = 2

$\therefore$ r = 2

substituting r = 2 in equation (3)

$\frac{u^2}{\bigg(2 + \frac{1}{2}\bigg)^2} + \frac{v^2}{\bigg(2 - \frac{1}{2}\bigg)^2} = 1 \\ \frac{u^2}{\frac{25}{4}} + \frac{v^2}{\frac{9}{4}} = 1.....(4)$

Therefore, circle with |z| = 2 maps into ellipse in w-plane with

$a = \frac{5}{2}, b = \frac{3}{4}$ & $e^2 = 1 - \frac{b^2}{a^2} = 1 - (9/25) = 16/25$

$\therefore e^2 = \bigg(\frac{10}{25}\bigg)$ i.e.

$e = \bigg(\frac{4}{5}\bigg) = 0.8 \\ \therefore e \lt 1 (ellipse)$

Equation (4) can be written as $\frac{u^2}{25} + \frac{y^2}{9} = \frac{1}{4}$