Conic Sections formulae using Complex Numbers

Question

I have been preparing for the JEE Examination here in India and have been studying Complex Numbers for the past few days. One of the topics which falls under Complex Numbers is their application in Coordinate Geometry (Conic Sections). These include the following:

Equation of lines in different forms (parametric/non-parametric)$$$$ Equations of Circles with various conditions (for example centered at $z_0$, or orthogonal to another circle and so on)$$$$ Equations of Ellipses and Hyperbolas and so on.

$$$$ There are several other applications/equations mentioned which I haven't written here. The trouble I'm facing is that the book I use for Complex Numbers (Algebra for JEE Main and Advanced, by SK Goyal) only has the formulae listed without showing the derivations. I find it extremely hard to just accept and use a formula/result without knowing how it came into existence.

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I would be grateful if somebody could please mention a source/reference book from which I could actually learn how to derive/reach all the formulae used for represinting Conics using Complex Numbers. The book does not necessarily have to match the level of the JEE Advanced Examination; it can be higher than that too. However I would prefer it if the book was at the level suitable for the JEE Advanced level only.

$$$$Many thanks in advance! $$$$ Edit:$$$$ The results mentioned in my book are as follows: $$$$

$1) $The equation of the line joining $z_1$ and $z_2$ is $$z(\bar{z_1}-\bar{z_2})-\bar{z}(z_1-z_2)+ z_1\bar z_2-z_2\bar z_1=0 \text{ (non parametric form).}$$

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$2) $Three points are collinear if $$\begin{vmatrix}z_1&\bar{z_1}&1\\z_2&\bar {z_2}& 1\\z_3&\bar {z_3}& 1\end{vmatrix}=0$$

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$3)$$\bar{a}z+\bar z a+b=0$ where $b\in \mathbb R$ describes the equation of a straight line (I don't know what $a$ is, nor what $\iota b$ is).

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$4)$ The real and complex slope (I don't know what is meant by 'real' and 'complex') of the line $\bar{a}z+\bar z a+b=0$ are $-\dfrac{\Re(a)}{\Im(a)}$ and $-\dfrac{a}{\bar a}$ where $b\in \mathbb R$.

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$5) $If the lines $\bar{a}z+\bar z a+k_1=0$ and $\bar{b}z+\bar b a+k_2=0$ $k_1,k_2\in \mathbb R$ are perpendicular to each other, then $$\bar{a}b+\bar ba=0$$

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$6)$ $z\bar z +a\bar z +\bar az+k=0 $ where $k\in \mathbb R$ represents a circle with center $-a$ and radius $\sqrt{|a|^2-k}$.

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$7)$ If $|z-z_1|+|z-z_2|=2a$ where $2a>|z_1-z_2|$ then $ z $ represents an ellipse with foci at $z_1 \text{ and }z_2$ and $a\in \mathbb R^+ $.

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$8)$ If $|z-z_1|-|z-z_2|=2a$ where $2a<|z_1-z_2|$ then $z$ represents a hyperbola with foci at $z_1$ and $z_2$.

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$9)$ Equation of all circles orthogonal to $|z-z_1|=r_1\text{ and }|z-z_2|=r_2$ is (nothing further is mentioned).

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$10)$ $\left |\dfrac{z-z_1}{z-z_2}\right | = k$ is a circle if $k\neq 1$ and will represent a line if $k=1$.

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$11)$ The equation $|z-z_1|^2+|z-z_2|^2=k$ will represent a circle if $k\geq \frac12 |z_1-z_2|^2$.

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$12)$ If $\arg\left(\dfrac{(z_2-z_3)(z_1-z_4)}{(z_1-z_3)(z_2-z_4)}\right)=0, \pm \pi$, then the points $z_1,z_2,z_3,z_4$ are concyclic.

Answer 1

1)Follows from 3. In 3) $a$ represents a direction (actually normal to a direction), and $b$ represents a distance from the origin -- or $\left |\frac {b}{a}\right|$ does.

2)In cartesian:

$\left|\begin {array} {} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{array}\right| = 0$ suggests $(x_1,y_1), (x_2,y_2), (x_3,y_3)$ are colinear

-- or the pyramid formed by the 3 points in the plane $z = 1$ together with the origin forms a pyramid of zero volume.

$\begin {bmatrix} {} z_1&\bar z_1&1\\z_2&\bar z_2&1\\z_3&\bar z_3&1\end{bmatrix} = \begin {bmatrix} x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{bmatrix}\begin {bmatrix} 1&1&0\\i&-i&0\\0&0&1\end{bmatrix}$

The right most matrix is non-sigular.

6)$\overline{(z-z_1)}(z-z_2) = k$ is the locus of points $z$ such that $\angle z_1zz_2$ is constant.

7)is the very definition of an ellipse. The distance from one focus to a point on the ellipse to the other focus is constant.

10)$|z - z_1| = k|z-z_2|\\ (z - z_1)\overline (z-z_1) = k^2 (z - z_2)\overline (z-z_2)\\ x^2 + y^2 + 2xx_1 + 2yy_1 + x_1^2 + y_1^2 = k^2(x^2 + y^2 + 2xx_2 + 2yy_2 + x_2^2 + y_2^2)\\ (k^2 - 1)(x^2 - a)^2 + (k^2 - 1)(y^2 - b)^2 = r^2$

Update 2025

11. $|z-z_1|^2 + |z-z_2|^2 = k$ if $k>\frac 12 |z_1-z_2|^2

$(x - x_1)^2 + (y-y_1)^2 + (x-x_2)^2 + (y-y_2)^2 = k$
$2x^2 - 2x(x_1+x_2) + 2y^2 + 2y(y_1+y_2) + x_1^2+x_2^2+y_1^2+y_2^2 = k$
$2(x^2 - \frac {x_1+x_2}{2})^2 + 2(y^2 - \frac {y_1+y_2}{2})^2 = k - \frac 12 ((x_1-x_2)^2 + (y_1-y_2)^2$

This is a circle if $k- \frac 12 |z_1-z_2| > 0$

For some intuition:

The parallelogram rule: The sum of the squares of the diagonal of a parallelogram equals the sum of the squares of the sides

That is, for parallelogram $ABCD, AC^2 + BD^2 = AB^2+BC^2+CD^2+DA^2 = 2AB^2 + 2BC^2$

If we construct our parallelogram with vertexes at $a,b,z$ we can find $z’ = a+b-z$ that completes the parallelogram

$2|z-a|^2 + 2|z-b|^2 = |a-b|^2 + |z-z’|^2$

$|a-b|^2 + |z-z’|^2 = 2k$ in the original statement of the problem.

$|z-z’|$ is constant.

Consider now the set of all parallelograms with one diagonal fixed and the other of fixed lenght.

Answer 2

There's detail discussion for general conics in Masayo Fujimuraa's paper:

The Complex Geometry of Blaschke Products of Degree 3 and Associated Ellipses