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http://www.wolframalpha.com/input/?i=5+%2B+%286.7a%29%2B+%28a+%5E+0.8%29

What does "imaginary part" mean?

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Imaginary numbers are those which are defined as the following:

$$\sqrt{-1} = i$$

In yours it says $0i$ which just evaluates to $0$, essentially there is no imaginary part.

Wolfram Alpha is putting it in a form called Complex Numbers. These have an imaginary and real part, and are defined as: $a+bi$. If the imaginary part is equal to $0$ then you are essentially left with a real number.

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  • $\begingroup$ Thanks Link. I will accept it as soon as possible. (8 mins left) $\endgroup$ – user117031 Dec 20 '13 at 20:23
  • $\begingroup$ @user117031 No problem! Glad to be of help. $\endgroup$ – Rivasa Dec 20 '13 at 20:26
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Some equations, like $x^2+1=0$ for instance, have no solutions in the real numbers. You may have heard things that allude to this in your education so far—perhaps you've heard it said that the polynomial $x^2+1$ has no real roots, perhaps implying that it may have solutions that are "not real". The same types of equations may have solutions in a different number system called the complex numbers. The complex numbers are a number system in which negative numbers have square roots, and often they are defined in terms of a number $i=\sqrt{-1}$ known as the imaginary constant. We may then write any complex number in the form $a+bi$ where $a$ and $b$ are real numbers, and we say that the number $z=a+bi$ has real part $a$ and imaginary part $b$. So, for example, $i$ has real part $0$ and imaginary part $1$, and is a solution of the equation I mentioned at the start:$$i^2+1=-1+1=0$$

So when you give Wolfram Alpha certain types of polynomials or equations, it checks for solutions in the complex numbers as well as in the real numbers since often there will be complex solutions but not real solutions to a given equation.

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