# Complex zeros

1. Aug 3, 2011

### vrc

hey,

I was solving the following equation: f(x) = 3x^2+7x+10
this gives me the following complex zeros:

-1.17+1.404i
-1.17-1.404i

I rewrite those as

-(1.17-1.4041i)
-(1.17+1.404i)

How do I visualize this complex zeros, because like f(x)=x^2-2x+1=(x-1)^2
that is the multiplication between 2 functions (x-1)
is there some kind of relation between those interpretations ?

thank you !

2. Aug 3, 2011

### gb7nash

A nice fact to know is that a is a root of f(x) if and only if (x-a) is a factor of f(x). So, if a polynomial f(x) has roots a1, a2, ... , an, and leading coefficient b, we can write f(x) as:

f(x) = b(x-a1)(x-a2)...(x-an).

In the case of f(x)=x^2-2x+1, x = 1 is a double root, so we can write f(x) as:

f(x) = (x-1)(x-1) = (x-1)2

Knowing this, how do we rewrite f(x) for your problem?

3. Aug 4, 2011

### vrc

well I had already calculated that in my course:

f(x) = 3x^2+7x+10=3*(x+(1.17-1.404i))*(x+(1.171.404i))

it's the last therm I want to have a visual image of like the example I have given:

f(x) = x^2-1 = (x-1)*(x+1) , so g(x) = x-1 multiply with h(x) = x-1 gives (fx) = (x-1)^2 = x^2-2x+1

I hope I answered your question ?

thank you

grtz

4. Aug 4, 2011

### HallsofIvy

Staff Emeritus
Yes,
$$3x^2+ 7x+ 10= 3(x+(1.17-1.404i))(x+(1.17+ 1.404i))$$
$$= 3((x+ 1.17)- 1.404i)((x+ 1.17)+ 1.404i)$$
$$= 3((x+ 1.17)^2+ (1.404)^2)$$

5. Aug 4, 2011

### vrc

yes that true, but so far I nobody still give me an answer how I have to visualize the complex zeros

3*(x+(1.17-1.404i))*(x+(1.171.404i))

I can't put this into a plotter because it's not on the xy plane, but on the complex plane isn't a variable x...
so I'm still questioning

thanks

grtz

6. Aug 4, 2011

### HallsofIvy

Staff Emeritus
It is not at all clear to me to what you mean by "visualizing" the complex roots. If you mean in the sense that you can see real roots as where the graph crosses the x-axis, you might prefer this: If a quadratic equation has complex roots then its graph will not cross the x-axis but if its vertex is at $(x_0, y_0)$ then its complex roots are $x_0\pm i \sqrt{y_0}/a$ where "a" is the leading coefficient.

That can be seen from the form I gave before: If a quadratic equation has leading coefficient a and roots $x_0\pm y_0i$, then it can be written as
$$a(x- (x_0+ iy_0))(x- (x_0- iy_0)))= a((x- x_0)^2+ y_0^2)$$

when $x= x_0$, $y= ay_0^2$.

7. Aug 4, 2011

### vrc

thank you for your explenation, now it makes more sense to me !

so I can pressume because that such an equation is rewritable in the form you wrote
and that those zeros are situated around the vertex of the equation
because it isn't real those zeros are then complex

something like this ?

thanks

grtz

8. Aug 6, 2011

### jackmell

Assuming you have some experience in complex variables and know that a complex-valued function of a complex variable has a real part and an imaginary part, both of which are surfaces in 3D, this Wolfram demonstration might help you "visualize" complex zeros:

http://demonstrations.wolfram.com/LocationOfComplexRootsOfARealQuadratic/

(place the cursor over the plot, hold down the left mouse button to rotate the plot interactively in order to see it better however I'm not sure if you need Mathematica ver 8 and the associated browser add-on's to get this interaction)

Last edited: Aug 6, 2011
9. Aug 6, 2011

### vrc

very good post !
I have to add an extra dimension to de xy plane, it's like a 3d cartesian plane but on the z plane (assuming the y is vertical and x is horizontal) the imaginary multiples of i.

z^2-4z = (x+iy)^2-4(x+iy) = x^2+y^2-4x-2iy,
the real part of this solution is represented by the Re-axle = funttion of x and y, the imaginary solution can be found at the y axle

is this correct ?

thank you

10. Aug 6, 2011

### jackmell

Ok, that is really, really confussing and a "static" plot of it like you posted is virtually useless because you can't really see what's going on without having the ability to rotate it interactively.

Let me try to help you. Consider the real function f(x)=x^2-4x. That is really, only a "slice" of the more general complex function f(z)=z^2-4z in which the imaginary part of z=x+iy is simply set to zero. Now consider the more general case by letting the complex variable z=x+iy. Then we have:

$$f(z)=z^2-4z$$

So that in terms of the variables x and y we can write:

$$f(z)=u(x,y)=(x+iy)^2-4(x+iy)$$

$$u(x,y)=x^2+2xyi-y^2-4x-4iy$$

$$u(x,y)=x^2-y^2-4x+i(2xy-4y)$$

So that the real and imaginary components of f(z) are then:

$$Re(f)=x^2-y^2-4x=g(x,y)$$

$$Im(f)=2xy-4y=h(x,y)$$

Ok, then g(x,y) and h(x,y) are really functions of two variables x and y and these functions are surfaces in 3D and that demonstration is then plotting g(x,y) and h(x,y) and then superimposing the "complex zeros" on them where the functions g(x,y) and h(x,y) are zero at the same time.

Last edited: Aug 6, 2011
11. Aug 6, 2011

### lavinia

A classic picture of the zeros of a complex polynomial comes from viewing is as describing an incompressible fluid that flows along the the lines, real part = constant. The zeros of the function are the points where the velocity of the flow is zero.

You can find pictures of this in many books.

12. Aug 6, 2011

### vrc

these are the answers I looked for !
no offence to other replies!

so I can assume every real 2d function is a part of a greater 3d complex function where the imaginary part is defined zero ...

somehting like this ?

thank you !

grtz

13. Aug 7, 2011

### jackmell

Close enough: Every function f(x,y) of two variables, including functions such as f(x,y)=3, f(x,y)=2x and the likes, is the real part of some complex function h(z)=h(x,y)=f(x,y)+ig(x,y) and when g(x,y)=0, then h(z)=f(x,y)=a real function.

14. Aug 7, 2011

### vrc

and that is why it's called imaginary because it's still there but for real functions it has the value 0.

now this alle make's more and more sense, I can calculate with complex numbers and function, but hadn't no clue what actually happens in a graph

htank you so much

if you have even more (good) information like pdf's etc.., I'd like to have those !

grtz

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