# Is it possible to convert a complex solution to a real solution? [closed]

I am a beginner, when it comes to higher level math, so bear with me, But I have been learning about imaginary numbers, and I found something quite concerning.

Imaginary numbers are used to resolve negative radicals. I learned that imaginary numbers stay in the equation, once they are added and no matter how much you simplify an equation, you cannot get rid of the imaginary number, i.

I see the reason for this, as it is the only way to represent the true nature of the equation, i is used in the real world, specifically for fields such as Electricity, Spectrum Analyzers, and the Mandelbrot Set. However, at some point, the value of your solution must have some concrete value, right?

My question: Is there ever a time, when you can "simplify out" i? Or in other words, convert the complex solution into a real solution? If not, how is the equation handled, if you can never find the "concrete" value of a solution?

• You don't because you can't (with regards to the last question) Sep 2, 2022 at 16:10
• I think some programming languages allow complex numbers as a data type. Sep 2, 2022 at 16:11
• You can't find a "tangible" solution to an equation without solutions in $\Bbb R$. Please don't think "imaginary" numbers are meaningfully more or less "real" than "real numbers" Sep 2, 2022 at 16:13
• Do you mean the real part and the imaginary part of a complex number ? Those are actually real numbers. But I have difficulties to grasp what is really asked. Sep 2, 2022 at 16:13
• The same way that the arithmetic of real number is arithmetic on points of a line, the arithmetic of complex numbers is arithmetic on points of a plane. When you do operations the point moves around. The $i$ can be eliminated from the result, if this ended up in the real line and if you have symbols in your language to represent this real number.
– plop
Sep 2, 2022 at 16:18

No. It is not possible to simply rearrange things so that you turn an imaginary solution into a real solution.

Why? Because real solutions are where our function crosses the x-axis, and "making" those complex solutions "real" would involve adding more crossings to our graph (which would be changing the problem entirely not just simplifying it or rearranging it).

Let's say the function is $$f(x)=x^2+bx+c$$, with the complex roots $$x_{1,2}=-\frac b2\pm i\frac{\sqrt{-(b^2-4c)}}2$$. Then you can transform $$f(x)$$ to $$g(x)=-x^2-bx-\frac{b^2}2+c$$. This is the reflected function $$f(x)$$ at the apex. The roots of $$g(x)$$ then are $$x_{1,2}=-\frac {b}{2}\pm\frac{\sqrt{-(b^2-4c)}}2$$, which are real.

Remark

The answer refers to the original question with a quadratic function:"How would I simplify down from here to find the 2 "real" values of X?"

The answer to your question is no. This is because any real number can be seen as (a particular case of) a complex number and not the other way around. Indeed, any real number $$x$$ is a particular case of the complex number $$x+yi$$, with $$y=0$$.

If something has an imaginary part, then you can never make that thing real. The expression $$3 + \sqrt{-2}$$ never "becomes" a real number.

However, there are methods that make use of complex numbers to make problems involving real-valued functions easier. Two ways we can do this are:

1. Extending the real function into the complex plane. This lets us apply complex methods (which, ironically, are often simpler than the ones you need to use if you stay in the real numbers), and often the solutions we get can be reduced back down to deal with just the real numbers once we're done. For example, in the complex field the exponential function $$e^z$$ and the trigonometric functions $$\cos$$ and $$\sin$$ are innately related, so you can derive a lot of trigonometric identities by messing around with exponentials instead.

2. In a similar vein, there are two functions - $$\Re(z)$$ and $$\Im(z)$$ that return the "purely real" and "purely imaginary" parts of a complex number. For example, $$\Re(3 + 2i) = 3$$ and $$\Im(3 + 2i) = 2$$. Sometimes, rather than extending our real-valued function to the complex plane, we instead put it in as the real part of a complex-valued function, do all the work with this new function, and then at the end extract our solution with the $$\Re(z)$$ function. We might even have two functions that are tightly tied to each other such that it makes sense to combine them as $$f(x) + i g(x)$$, or the like. This is roughly what happens in electrical engineering where, again, it's easier to work in complex exponential functions than real trigonometric ones.

In the end, it's all a trick to make things easier, just like you might prove something about a sequence of integers by working with a function of real numbers instead.