Quadratic equation with complex numbers in 3D

Hello! Since school, solving quadratic equations (KU), for example

x^{2} + x + 1 = 0

$x ^ 2 + x + 1 = 0$ , got roots with an imaginary component,

x = - f r a c 12 p m i f r a c s q r t 3 2

$x = - \ frac {1} {2} \ pm i \ frac {\ sqrt3} {2}$ and if you want to see how the graph crosses the axis

Y

$Y$ in points

x = - f r a c 12 p m i f r a c s q r t 3 2

$x = - \ frac {1} {2} \ pm i \ frac {\ sqrt3} {2}$ , I found on the Internet graphics like:

How does the graph with the imaginary part look (in my thoughts) in 3D (

X b o t Y b o t I

$X \ bot Y \ bot I$ ), and is the topic of this article.

PS: Under the cut heavy animations

As usual, the graph of f-ktsii consists of points, and points are under construction on intersection of axes

X

$X$ and

Y

$Y$ .
The schedule of f-tsii with a complex component

Y_{c o m p l e x} = F (X_{c o m p l e x})

$Y_ {complex} = F (X_ {complex})$ ,
')
Where

X_{c o m p l e x} = X_{r e} + X_{i m} =

$X_ {complex} = X_ {re} + X_ {im} =$

b e g i n b m a t r i x X_{r e} X_{i m} e n d b m a t r i x

$\ begin {bmatrix} X_ {re} \\ X_ {im} \ end {bmatrix}$ - vector

Y_{c o m p l e x} = Y_{r e} + Y_{i m} = b e g i n b m a t r i x Y_{r e} Y_{i m} e n d b m a t r i x

$Y_ {complex} = Y_ {re} + Y_ {im} = \ begin {bmatrix} Y_ {re} \\ Y_ {im} \ end {bmatrix}$ - vector

X_{c o m p l e x}

$X_ {complex}$ can be represented as a 3-dimensional vector

b e g i n b m a t r i x X_{r e} 0 X_{i m} e n d b m a t r i x

$\ begin {bmatrix} X_ {re} \\ 0 \\ X_ {im} \ end {bmatrix}$

[X_{r e}] - X a x i s

$[X_ {re}] - X axis$

[0] - Y a x i s

$[0] - Y axis$

[X_{i m}] - a x i s I

$[X_ {im}] - axis I$
Similarly

Y_{c o m p l e x} b e g i n b m a t r i x 0 Y_{r e} Y_{i m} e n d b m a t r i x

$Y_ {complex} \ begin {bmatrix} 0 \\ Y_ {re} \\ Y_ {im} \ end {bmatrix}$

[0] - a x i s X

$[0] - axis X$

[Y_{r e}] - Y a x i s

$[Y_ {re}] - Y axis$

[Y_{i m}] - a x i s I

$[Y_ {im}] - axis I$

Intersection point

X_{c o m p l e x}

$X_ {complex}$ and

Y_{c o m p l e x}

$Y_ {complex}$ will be equal to the sum of the vectors

X_{c o m p l e x}

$X_ {complex}$ and

Y_{c o m p l e x}

$Y_ {complex}$

b e g i n b m a t r i x X_{r e} 0 X_{i m} e n d b m a t r i x

$\ begin {bmatrix} X_ {re} \\ 0 \\ X_ {im} \ end {bmatrix}$ +

b e g i n b m a t r i x 0 Y_{r e} Y_{i m} e n d b m a t r i x

$\ begin {bmatrix} 0 \\ Y_ {re} \\ Y_ {im} \ end {bmatrix}$ =

b e g i n b m a t r i x X_{r e} Y_{r e} X_{i m} + Y_{i m} e n d b m a t r i x

$\ begin {bmatrix} X_ {re} \\ Y_ {re} \\ X_ {im} + Y_ {im} \ end {bmatrix}$

With the intersection figured out.

Next to build the graph you need to decide on the change

X_{r e}

$X_ {re}$ and

X_{i m}

$X_ {im}$ along the axis

X_{c o m p l e x}

$X_ {complex}$ , for this you need root KU. There are two options:

To make $X_ {im}$ constant and change only $X_ {re}$ from the root of KU;
Get the angle between $X_ {re}$ and $X_ {im}$ from the root ku and move along $X_ {complex}$ building up $X_ {re}$ , $X_ {im}$ calculate taking into account the angle and $X_ {re}$ .