What is the Nature of the Solution to x^2 = 1?

AI Thread Summary
The equation x^2 = 1 has solutions x = ±1, which are both real numbers. While the book categorizes these solutions as complex, this can be misleading since complex numbers typically include an imaginary component. In this case, the solutions can be expressed as complex numbers with zero imaginary parts (b=0), highlighting that all real numbers are also complex numbers. However, the distinction can create confusion, as many would simply classify 1 as a real number. Ultimately, the discussion emphasizes the relationship between real and complex numbers while acknowledging the potential for misunderstanding in terminology.
robertjford80
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Homework Statement



solve the following and determine if it is complex, real and/or imaginary

x2 = 1



The Attempt at a Solution



The answer is x = +-1 which I agree with but the book says that that is complex and real. How complex? Complex numbers have the form a + bi, there's no bi in that answer.
 
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##\mathbb{R} \subseteq \mathbb{C}##

A complex number ##c## can represented as ##c = a + bi## where ##a,b \in \mathbb{R}##

In this case, b=0. A purely imaginary number is when a=0.
 
robertjford80 said:

Homework Statement



solve the following and determine if it is complex, real and/or imaginary

x2 = 1



The Attempt at a Solution



The answer is x = +-1 which I agree with but the book says that that is complex and real. How complex? Complex numbers have the form a + bi, there's no bi in that answer.

They're trying to get you to see that real numbers are complex numbers. But it's a little misleading, since clearly if someone asks if 1 is "real or complex," the answer is real ... even though technically, the right answer is "both." I think this is a bit of a trick question.
 
thanks, I did feel tricked, but I get the point now.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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