What is the reason for this ?
$ (i^4)^{\frac 12} = 1^{\frac 12} = 1$
$(i^4)^{\frac 12} = {({(i^2)}^2)}^{\frac 12} = i^2 = -1 $
Here, $$ i=\sqrt{-1}$$
Asked
Active
Viewed 125 times
3
-
How do you define a square root on complex numbers? Can't just take the positive. – Kaynex Apr 06 '18 at 16:25
-
You need to be very careful with fractional powers of complex numbers. With the more familiar real numbers, you need to remember that there may be two solutions to $x^2 = y$. This is also true of the complex numbers with the added complication that there is not such a neat way to pick the primary solution. – badjohn Apr 06 '18 at 16:25
-
THIS ANSWER, THIS ANSWER, and THIS ONE might be helpful to understand the underlying issues. – Mark Viola Apr 06 '18 at 17:52
3 Answers
2
The reason for that is that every complex number (excepto $0$) has two square roots. The notation $a^{\frac12}$ is ambiguous.
José Carlos Santos
- 427,504
1
The properties of the powers that you are using are not valid for complex numbers if the exponents are not integer.
Emilio Novati
- 62,675
-
Thanks for this answer. The invalidity of the rule $(x^a)b=x^{ab}$ when $x$ is not a positive real and the exponents are not integers needs perpetual repetition. – Lubin Apr 06 '18 at 16:40
0
Because of the discontinuous nature of the square root function in the complex plane, in general it is not true that $\sqrt{zw}=\sqrt{z}\sqrt{w}$, but if $z,w$ nonnegative real numbers it becomes true. Assuming this rule you can "proof" that 1=-1: $$-1=ii=\sqrt{-1}\sqrt{-1}=(\neq)\sqrt{(-1)(-1)}=\sqrt{1}=1$$
D.G
- 93
-
Actually, it IS true that $\sqrt{zw}=\sqrt{z}\sqrt{w}$ in the sense of set equivalence. That is to say that any value of $\sqrt{zw}$ can be expressed as the product of some value of $\sqrt{z}$ and some value of $\sqrt{w}$. Conversely, the product of any value of $\sqrt{z}$ and any value of $\sqrt{w}$ can be expressed as some value of $\sqrt{zw}$. – Mark Viola Apr 06 '18 at 17:50
-
-