The problem is those morons haven’t taken any of the advanced classes and probably got D’s in the basic ones.
this post was submitted on 21 Aug 2025
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Sqrt(-1) is still wrong tho. I'm commuting a sin by writting it. Correct expression is i^2=-1
Wolfram tells me sqrt(-1) = i and it hasn't lied to me yet.
In what meaningful way is i^2 = -1 different from sqrt(-1) = i?
sqrt(-1) = ±i. The negative answer is also valid.
Ah, good point; I'd forgotten that part.
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Wouldn't the square root just give plus/minus i? Seems correct enough.
No. The symbol √ signifies the principal square root of a number. Therefore, √x is always positive. The two roots of x, however, are ±√x. If you therefore have y²=x and you want to find y, you mustn't write y=√x, but rather y=±√x to be formally correct.
But (-i)^2=-1 as well. So we still need a convention to distinguish i from -i.
That's fairly simple: we restrict the complex phase to the range (-pi, pi] and the principal square root halves the complex phase. -1 has the phase value pi, so the principal square root has the the complex phase pi/2, so it's i, while -i has a phase of -pi/2
They’re the same thing. You just take the square root of both sides to get i = sqrt(-1).
Indeed, usually you would want to avoid a notation of sqrt(-1) or (-1)^(1/2). You would use e^(1/2 log(-1)) instead because mathematicians have already decided on a "natural" way to define the logarithm of complex numbers. The problem here lies with choosing a branch of the logarithm as e^z = x has infinitely many complex solutions z. Mathematicians have already decided on a default branch of the logarithm you would usually use. This matters because depending on the branch you choose sqrt(-1) either gives i or -i. A square-root is usually defined to only give the positive solution (if it had multiple values it wouldn't fit the definition of a function anymore) but on the complex plane there isn't really a "positive" direction. You would have to choose that first to make sure sqrt is defined as a function and you do that via the logarithm branch.
So, just writing sqrt(-1) leaves ambiguity as you could either define it to give i or -i but writing e^(1/2 log(-1)) then everyone would just assume you use the default logarithm branch and the solution is i.
Nah, sqrt(x) is the principal branch (the one with a positive real part) of x^½, and you can do (-1)^½ because it's just exponentiation.
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Always wear your glasses. Sans glasses, I read the Advanced Math panel saying the square root of -1=1, and thought, "that's doesn't sound right."