What's the thing with \sqrt{-1}=i

Answered question

2022-01-17

Whats

Answer & Explanation

nick1337

nick1337

Expert2022-01-18Added 777 answers

Step 1
We use the value of i=1 so that we can work more freely. For example consider the Quadratic Equation
x2+x+1=0
The discriminant for this is D=b24ac=3. So the roots have the value
x=1±3i2
which looks better when written with an i notation. That's all.
I don't know how this is taught in the US but to me, i encountered this when i was at high school, learning how to solve for Quadratic Equations when the Discriminant is less than 0.
Next, note that C doesn't have the same ordering as R. That is for any 2 real numbers a,b we have either:
a>b
a<b
a=b
But for complex number's this is not true. Since if you take i and 0, we must have either i>0 or i<0, but this isn't true.

Vasquez

Vasquez

Expert2022-01-18Added 669 answers

Step 1 This seems to be very little different than the standard notation for 4=2 as opposed to 4=2. By convention, we have a concept of a principal branch of the square root. More importantly, though, is that there is no real difference between letting i: =1 and -i: =1 (though there is an imaginary difference!). One really needs only to notate the two roots of 1 by ±i. In addition, I first learned about imaginary numbers in a US school and we learned that i was s.t. i2=1 as well. As far as I can recall, the first important distinction comes when we consider how to represent the complex plane graphically and through the exponential series - but again, these are merely notational inconveniences rather than big problems.
alenahelenash

alenahelenash

Expert2022-01-24Added 556 answers

The proper way to define i is to define the laws for complex addition and multiplication the usual way using ordered pairs of real numbers, and then define i to be (0, 1). This distinguishes it from i which is (0, 1). One can then set the root of 1 to be i, but of course this last step is pretty arbitrary. You can see it either as wrong or as not wrong if you want to work with the last step as an arbitrary definition. The important thing is that we have an unambiguous definition of i using the initial steps specified above.

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