I Do complex roots have a physical representation on a curve?

[mohammed El-Kady]

If we have y=x^2 -4. This is represented by curve intersect x-axis at (-2, 0) and (2, 0) or if we wish to find it algebraically we set y =0 then we solve it. The roots must lie on the curve.
when y=x^2+4 the roots are 2i and -2i "complex" consequently there is no intersection with x-axis, so where the points (2i, 0) and (-2i, 0) on the curve?
Does it have another dimension or imagine dim.? if yes, is that mean the curve is solid?

 

[fresh_42]

If we have y=x^2 -4. This is represented by curve intersect x-axis at (-2, 0) and (2, 0) or if we wish to find it algebraically we set y =0 then we solve it. The roots must lie on the curve.
when y=x^2+4 the roots are 2i and -2i "complex" consequently there is no intersection with x-axis, so where the points (2i, 0) and (-2i, 0) on the curve?
Does it have another dimension or imagine dim.? if yes, is that mean the curve is solid?

You'll have to say what you are talking about. On the one hand you talked about the graphs $\{\,(x,y)\in \mathbb{R}^2\,|\,y=x^2\pm 4\,\}$, which are parabolas in the Euclidean plane, one which crosses the $x-$axis twice, and one which does not.

Now you brought in complex numbers. That is another problem, namely how real polynomials in one indeterminate split over the complex numbers. This is an algebraic theorem and pure abstract algebra. However, you took this now complex situation and asked about its real correspondence? There is none. The inclusion $\mathbb{R} \subsetneq \mathbb{C}$ is a strict one. If you consider the complex numbers as a two dimensional real vector space, then the new function $f\, : \, \mathbb{C} \longrightarrow \mathbb{C}$ with $f(z)=z^2+4$ is a different one, and its graph has four real dimensions, making it difficult to draw.

 

[mohammed El-Kady]

You'll have to say what you are talking about. On the one hand you talked about the graphs $\{\,(x,y)\in \mathbb{R}^2\,|\,y=x^2\pm 4\,\}$, which are parabolas in the Euclidean plane, one which crosses the $x-$axis twice, and one which does not.

Now you brought in complex numbers. That is another problem, namely how real polynomials in one indeterminate split over the complex numbers. This is an algebraic theorem and pure abstract algebra. However, you took this now complex situation and asked about its real correspondence? There is none. The inclusion $\mathbb{R} \subsetneq \mathbb{C}$ is a strict one. If you consider the complex numbers as a two dimensional real vector space, then the new function $f\, : \, \mathbb{C} \longrightarrow \mathbb{C}$ with $f(z)=z^2+4$ is a different one, and its graph has four real dimensions, making it difficult to draw.

thanks for your explanation i just need to know where the roots exists on the curve

 

[fresh_42]

The imaginary roots do not exist on the real curve. They exist in the complex hyperplane, if the function is considered to be complex, in which case the "curve" is a there dimensional hyperspace of a four dimensional space.

The roots "exist" only in the complex extension of the real numbers, as algebraic entities in $\mathbb{R}[x]/\langle x^2+1 \rangle$

 

[mfb]

If you plot $|x^4+4|$ for x in the complex plane then you'll see it is 0 at +2i and -2i. This is not in the real line as the roots are not real. They are complex.
WolframAlpha can plot the real and imaginary part

 

[mohammed El-Kady]

If you plot $|x^4+4|$ for x in the complex plane then you'll see it is 0 at +2i and -2i. This is not in the real line as the roots are not real. They are complex.
WolframAlpha can plot the real and imaginary part

Thank you for the valuable graphs, its explains too much information for me

 

[WWGD]

The roots exist in the curve $f(z)=z^4+4$ . Note, as someone said, $x^4 \pm 4$ is a subspace of $z^4+4$. The map $f(z)$ goes from $\mathbb R^2 \rightarrow \mathbb R^2$.