# Fraction Degree of Complex Number [closed]

It is known that $i^{25}=i$, but ${i^{25}=(i^{4})^{\frac{25}{4}}}=1$. So, We have contradiction.

Let me rephrase it: what kind of limitation should be on multiplication of degrees?

What do you expect me to clarify? There are 3 well-known formulas 1. ${a^n} {a^m}=a^{n+m}$ 2. ${a^n}: {a^m}=a^{n-m}$ 3. ${(a^{m})^{n}=a^{mn}}$, But without limitation (for every $n$ and every $m$) they work only if $a$ is real and greater than zero. These formulas are also work in some other cases. My question: what are limitation on these formulas in case if $a=i$?

## closed as unclear what you're asking by user223391, Ofir Schnabel, Servaes, N. F. Taussig, Harish Chandra RajpootOct 27 '15 at 10:47

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• Here is a question you might want to consider: what exactly is the definition of $z^{25/4}$? – Paul Sinclair Oct 27 '15 at 3:38

There is no contradiction, only a misconception. When working with complex numbers, $1^{\frac{25}{4}}$ has multiple values, not just $1$. (Or, to put it more generally, real positive numbers to fractional exponents can return multiple complex values).

A correct manipulation would be $i^{25} = ({i^4})^6 \cdot i = 1^6 \cdot i = i$.

• @Dr.MV Thanks Mark, cheers. – Deepak Oct 27 '15 at 3:42
• You're welcome. My pleasure. - Mark – Mark Viola Oct 27 '15 at 3:43
• I know that for many years. It is exactly why I ask. Let me rephrase it: what kind of limitation should be on multiplication of degrees? – Victor Victorov Oct 27 '15 at 3:48

When dealing with complex functions, you often run into multi-valued functions, because when you circle around a singularity, it will pick up some constant value. There are three ways in which this is often handled:

1. You can restrict the function to domains, called branches, that do not completely circle a singularity.
2. You can live with the concept of a "multi-valued" function, treating the function value as being a set of complex numbers instead of a single complex number, or
3. You can build a layered extension of the complex numbers, called a Riemann surface, to act as the domain, so that the function picks up each of its different values on a different layer.

2. Multiplication of exponents always works, and your $i = 1$ example is not a problem because the actual value is a set that contains both.