This question already has an answer here:

I understand how to solve problems dealing with imaginary numbers, but I don't understand the reason why they exist and what they really do. Could somebody please explain to me what the point of them is? What I don't understand is that wouldn't multiplying by negative one just do the same thing?


marked as duplicate by Git Gud, hardmath, Mark Fantini, anomaly, Lucian Nov 9 '14 at 0:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ You are mistaken when you say "it takes out the negative". $\endgroup$ – Paul Sundheim Nov 9 '14 at 0:04
  • $\begingroup$ @PaulSundheim What would it do then? $\endgroup$ – IHeartBunnies Nov 9 '14 at 0:04
  • $\begingroup$ @IHeartBunnies They give an answer to "$\sqrt{-1}={}?$". Multiplying $-1$ by negative one wouldn't give you the right answer to your question. (Also, it turns out that imaginary and complex numbers are useful in other branches of math, like trigonometry. It can be shown that:$$(\cos(A)+i\sin(A))(\cos(B)+i\sin(B))=(\cos(A+B)+i\sin(A+B)),$$a fact that is very useful.) $\endgroup$ – Akiva Weinberger Nov 9 '14 at 0:19
  • $\begingroup$ Ok, thanks. That makes a little more sense now. $\endgroup$ – IHeartBunnies Nov 9 '14 at 0:21
  • $\begingroup$ Why does 14 exist? $\endgroup$ – Mariano Suárez-Álvarez Nov 9 '14 at 2:53

When you are asking about why they exist, I take it you mean why they were developed? Because if you're really asking about whether numbers exist, that becomes a philosophical and rather complicated question about our ontological commitments to mathematical entities.

They were first noticed possibly when mathematicians were solving quadratic polynomials, i.e. $ax^2+bx+c=0$. You'll quickly notice that sometimes we get solutions involving taking the square root of negative value. Mathematicians dismissed this as being absurd until they began to work on finding a formula for the roots of the general cubic polynomial, i.e. $ax^3+bx^2+cx+d=0$.

As for what they do, they have a lot of applications within and outside of mathematics. We're able to solve a lot of problems which appear to be firmly fixed in the real numbers using complex numbers. Within mathematics, this can be seen in geometry, calculus, etc. Outside of mathematics, it is extremely useful to physics and thus useful to engineering, particularly electrical engineering.

If our goal was to "get rid" of the negative, sure, multiplying by a negative number would get rid of it symbolically but then that changes our equation algebraically.

  • 3
    $\begingroup$ Just to be clear: "absurd" results in the quadratic formula always indicate non-real solutions. On the other hand, Cardano's formula for solving the cubic sometimes involves a root of a negative, but if we pretend that $i$ exists and can be manipulated normally, the simplified form of the formula is real (imaginary parts in different terms cancel). $\endgroup$ – JHance Nov 9 '14 at 0:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.