Solve over complex.

(2 - x)[sup]1/2[/sup] + 10 = 4

I was thinking it was no solution, but I'm not sure with complex numbers.

No solution.

If z is any complex number, and z^(1/2) = a + bi, then:
1) a > 0.
or:
2) a = 0 and b >= 0.

Proof (a trigonometric, not algebraic approach):
1. z is not a real number.
If z = r cis(theta)
Where -pi < theta < pi (*)
Then sqrt(z) = sqrt(r) cis(theta/2).
Therefore -pi/2 < arg(sqrt(z)) < pi/2.
Since Re(sqrt(z)) = sqrt(r) cos(arg(sqrt(z))), the cosine is always positive, meaning Re(sqrt(z)) > 0.

2. z is a real number.
In this case the proof is obvious (if z > 0 then a > 0, if z <= 0 then a = 0 and b >= 0).

In your example:
(2 - x)^(1/2) = -6

Even assuming 2 - x is a complex number, this cannot be solved since -6 doesn't satisfy either of the above mandatory requirements.

[hr]
(*) In z = r cis(theta) we can define either 0 <= theta < 2pi or -pi < theta <= pi. The latter is chosen to avoid ambiguity in fractional powers.

But if you were to define the theta range to include 2pi, it would have a solution... I picture roots as dividing the angle argument of the complex number in polar form, and for the resulting angle to be pi, the original number would have to have the angle 2pi..

So, there is math that currently can't be solved, that needs super-imaginary numbers? :P

[quote author=Adron link=board=36;threadid=7204;start=0#msg64722 date=1086998345]
and for the resulting angle to be pi, the original number would have to have the angle 2pi..
[/quote]
But that's not true. If the original number had the angle 2pi, then its argument would actually be 0, not 2pi. Because the argument is the angle modulo 2pi.
Then the resulting angle would be 0, not pi.
(In English, if the number is real and positive, its root is real and positive.)

There are too many reasons to restrict the argument to the range (-pi, pi] and not [0, 2pi).
For example, if z = x + yi, we may unambiguously define arg(z) = arctan(y / x).
(Even if x = 0... Then,
If y > 0, arg(z) = arctan(infinity) = pi/2
If y < 0, arg(z) = arctan(-infinity) = -pi/2)

And yes, there are quaternions and hypercomplexes and more classes of super-imaginary numbers. I don't know much about them, but I'm pretty sure this isn't what they solve.

Edit: To clarify, all the fundamental theorem of algebra guarantees is the solution in complex numbers of any polynomial equation. An equation of the form "z^(1/2) = a + bi" is not a polynomial equation, because the degrees of z are not integers. So yes, there are plenty of equations not solvable in complex numbers, and this is one of them.

[quote author=Yoni link=board=36;threadid=7204;start=0#msg64817 date=1087049521]
[quote author=Adron link=board=36;threadid=7204;start=0#msg64722 date=1086998345]
and for the resulting angle to be pi, the original number would have to have the angle 2pi..
[/quote]
But that's not true. If the original number had the angle 2pi, then its argument would actually be 0, not 2pi. Because the argument is the angle modulo 2pi.
Then the resulting angle would be 0, not pi.
(In English, if the number is real and positive, its root is real and positive.)
[/quote]

For the original number to be able to have the angle 2pi, the defined range for arguments would have to include 2pi, which was what lead to using a range such as 0 < argument <= 2pi...

Which then leads to completely counter-intuitive fractional power rules, since 1 = 1 cis(2pi), therefore one might think sqrt(1) = 1 cis(pi) = -1. It is a solution of z^2 = 1, but not the square root of 1, which has to be defined unambiguously.

Ah, you're right of course. It would lead to very strange things. I suppose we're better off with -pi to pi.

Good, I marked "no solution" on the test. I'm sure this will come up again some day.