# Thread: Multivalued functions. Can somebody explain?

1. ## Multivalued functions. Can somebody explain?

I really don't understand this:

The set of values of $\displaystyle log(i^{1/2})$ is $\displaystyle (n+1/4)\pi i$, with $\displaystyle n \in Z$, and that the same is true of $\displaystyle \frac{1}{2} log(i)$, but...

The set of values of $\displaystyle log(i^2)$ is not the same as the set of values of $\displaystyle 2 log(i)$.

2. Originally Posted by MathSucker
I really don't understand this:

The set of values of $\displaystyle log(i^{1/2})$ is $\displaystyle (n+1/4)\pi i$, with $\displaystyle n \in Z$, and that the same is true of $\displaystyle \frac{1}{2} log(i)$, but...

The set of values of $\displaystyle log(i^2)$ is not the same as the set of values of $\displaystyle 2 log(i)$.
I think that the point here is that the square root function, like the logarithm, is multivalued (or 2-valued in the case of the square root). So the set of values of $\displaystyle \log(z^{1/2})$ means the set of all the logarithms of both square roots of z.

But when you multiply $\displaystyle \log z$ by 2, you only get half the values of $\displaystyle \log(z^2)$ (the other half come from $\displaystyle 2\log(-z)$).