# Why Do Imaginary Numbers Exist [duplicate]

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, LucianNov 9 '14 at 0:19

• You are mistaken when you say "it takes out the negative". – Paul Sundheim Nov 9 '14 at 0:04
• @PaulSundheim What would it do then? – IHeartBunnies Nov 9 '14 at 0:04
• @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.) – Akiva Weinberger Nov 9 '14 at 0:19
• Ok, thanks. That makes a little more sense now. – IHeartBunnies Nov 9 '14 at 0:21
• Why does 14 exist? – Mariano Suárez-Álvarez Nov 9 '14 at 2:53

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. $ax^3+bx^2+cx+d=0$.
• 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). – JHance Nov 9 '14 at 0:19