# Finding the locus of a complex number?

## $z$ is a variable complex number such that $| z | = 1$ and $u = 3 z - \frac{1}{z}$. Show that the locus of the point in the Argand plane representing $u$ is an ellipse and find the equation of the ellipse.

Mar 14, 2017

${\left(\frac{X}{2}\right)}^{2} + {\left(\frac{Y}{4}\right)}^{2} = 1$

#### Explanation:

We have $z \overline{z} = {\left\mid z \right\mid}^{2} = {x}^{2} + {y}^{2} = 1$ This can be described also as

$z = {e}^{i t}$ for ${e}^{i t} = \cos \left(t\right) + i \sin \left(t\right) = x \left(t\right) + i y \left(t\right)$

Now

$u = 3 {e}^{i t} - {e}^{- \left(i t\right)} = 3 \left(\cos \left(t\right) + i \sin \left(t\right)\right) - \left(\cos \left(- t\right) + i \sin \left(- t\right)\right)$

or

$u = 3 \cos \left(t\right) - \cos \left(t\right) + i \left(3 \sin \left(t\right) + \sin \left(t\right)\right) = X + i Y$

or

$\left\{\begin{matrix}3 \cos \left(t\right) - \cos \left(t\right) = 2 \cos \left(t\right) = X \\ 3 \sin \left(t\right) + \sin \left(t\right) = 4 \sin \left(t\right) = Y\end{matrix}\right.$

so the ellipse equation is obtained as

${\cos}^{2} \left(t\right) + {\sin}^{2} \left(t\right) = {\left(\frac{X}{2}\right)}^{2} + {\left(\frac{Y}{4}\right)}^{2} = 1$

Mar 14, 2017

Loci is that of an ellipse with equation

${\left(\frac{x}{2}\right)}^{2} + {\left(\frac{y}{4}\right)}^{2} = 1$

#### Explanation:

Let $u = x + y i$, and $z = a + b i$; then

From $| z | = 1$ we have:

${a}^{2} + {b}^{2} = 1 \setminus \setminus \setminus \setminus \ldots . . \left[1\right]$

From $u = 3 z - \frac{1}{z}$ we have:

$x + y i = 3 \left(a + b i\right) - \frac{1}{a + b i}$
$\text{ } = 3 a + 3 b i - \frac{1}{a + b i} \cdot \frac{a - b i}{a - b i}$
$\text{ } = 3 a + 3 b i - \frac{a - b i}{{a}^{2} - {\left(b i\right)}^{2}}$
$\text{ } = 3 a + 3 b i - \frac{a - b i}{{a}^{2} + {b}^{2}}$
$\text{ } = 3 a + 3 b i - \left(a - b i\right) \setminus \setminus \setminus \setminus$ (from [1])
$\text{ } = 3 a + 3 b i - a + b i$
$\text{ } = 2 a + 4 b i$

And so:

$x = 2 a \implies a = \frac{x}{2}$
$y = 4 b \implies b = \frac{y}{4}$

${a}^{2} + {b}^{2} = {\left(\frac{x}{2}\right)}^{2} + {\left(\frac{y}{4}\right)}^{2}$
$\therefore {\left(\frac{x}{2}\right)}^{2} + {\left(\frac{y}{4}\right)}^{2} = 1$ (from [1])
Which is the equation f an ellipse with semi-minor axis $2$ and semi-major axis $4$