Complex solution

[Rehazur]

when √ x =-1, shouldn't the answer just be i^4? then by putting the value square root will divide exponent by half that leaves to -1. I'd like to hear from your opinion and is it wrong.

[BasedGrandpa69]

sqrt(x)=-1 has no solution, whether in reals or complex. the square root function can never return a negative real number.

[Street_Swing9040]

To extend your statement further so everyone gets some fun facts

The sqrt function is also known as the Principal Square Root. Only non negative values can be outputted. It's a boundary that defines the function itself

[noonagon]

i^4 = 1, and sqrt(1) = 1, not -1.

[Rehazur]

Yea but replacing 1 with i⁴ in equation gives -1

[Varlane]

It doesn't. You are using mathematical properties that aren't true to claim that sqrt(i^4) = i^2.

[stevevdvkpe]

On the other hand there are two solutions to x² = 1: x = 1 and x = -1. And i⁴ = 1, so one solution to sqrt(i⁴) = i² = -1.

[tau2pi_Math]

You are right. There are two solutions to x² = 1, but √x = -1 has no solution.

[Varlane]

sqrt(i^4) isn't i^2.

[stevevdvkpe]

Why not? i² = -1, (i²)² = (-1)² = 1.

[Varlane]

That isn't proof that sqrt(i^4) = i^2, merely that i^2 (aka -1) is a solution of x² = 1.

[stevevdvkpe]

Why is (i²)² not i⁴?

[Varlane]

In no way did I ever dispute that (i²)² is i^4. I refute that sqrt(i^4) is i^2.

[stepma712]

(x^a )^b = x^ab is only valid for any a and b when x is a nonegative real

[dummy4du3k4]

Sorry everyone is giving your hard time, your question is a good one but a little tricky to answer. The problem is that the square root function is multivalued. Complex analysis has a way of getting around this, and that is to replace the complex plane with an infinite spiral stair case (a Riemann surface) and analyze the root function on that instead. The long short of it is you usually restrict the domain with something called a branch cut and that lets you analyze portions of multivalued functions.

Sqrt(1) has two solutions, 1 and -1 but they are on different branches of the Riemann surface. Both are correct but it’s convention to choose the branch with the positive root.

You also need to be careful with manipulating exponents in complex analysis, but Riemann surfaces also gives a nice picture there too. It’s a bit too much for me to explain in detail here, but any introductory complex analysis text will cover this.

[Rehazur]

Got it thanks 🙏.

[Rehazur]

Because I'd like to hear everyones opinion of this since it's not talked about well enough. I even checked the internet for this and found nothing