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Under the transformation $w + zi = z + \frac{1}{z}$. Show that the map of the circle |z| = z is an ellipse in $w$-plane.
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Given $w + zi = z + \frac{1}{z}$......(1)

We know that $z = re^{i\theta}$

$\therefore$ and according to the given condition |z| = 2 = r

i.e. 2 = r $\therefore$ r = 2

Substituting $z = re^{i\theta}$ in equation (1)

$w + zi = re^{i\theta} + \frac{1}{re^{i\theta}}$ substituting r = 2 & $e^{i\theta} = cos\theta + isin\theta$

$w + zi = 2e^{i\theta} + \frac{1}{2} e^{-i\theta} \\ w + zi = 2cos\theta + 2isin\theta + \frac{1}{2}cos\theta - \frac{1}{2}isin\theta \\ w + zi = \bigg(2 + \frac{1}{2}\bigg)cos\theta + i\bigg(2 - \frac{1}{2}\bigg)sin\theta$

where $w = u + iv$

$u + iv + 2i = \frac{5}{2}cos\theta + i\frac{3}{2}sin\theta \\ u + i(v + 2) = \frac{5}{2}cos\theta + i\frac{3}{2}sin\theta$

Equating real and imaginary parts we get

$u = \frac{5}{2}cos\theta, (v + 2) = \frac{3}{2}sin\theta \\ \therefore cos\theta = \frac{2u}{5} , sin\theta = \frac{2(v + 2)}{3}$

Now in order to eliminate $\theta$ we will square the above two equations and add we get

$[(sin^2 \theta + cos^2 \theta) =1]$

$\therefore cos^2 \theta + sin^2 \theta = \frac{4u^2}{25} + \frac{4}{9}(v + 2)^2 \\ \therefore \frac{u^2}{\bigg(\frac{5}{2}\bigg)^2} + \frac{(v + 2)^2}{\bigg(\frac{3}{2}\bigg)^2} = 1$

Which is an equation of ellipse

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