# Complex exponential to a power

Say I have ##e^{2\pi i n}##, where ##n## is an integer. Then it's clear that ##(e^{2\pi i})^n = 1^n = 1##.
However, what if replace ##n## with a rational number ##r##? It seems that by the same reasoning we should have that ##e^{2\pi i r} = (e^{2\pi i})^r = 1^r = 1##. But what if ##r=1/2## for example? Then ##e^{2 \pi i (\frac{1}{2})} = e^{\pi i} = -1##. What am I doing wrong here?

• SACHIN KUMAR

## Answers and Replies

jedishrfu
Mentor
Isn't that Euler's identity you now have?

Add 1 to both sides to get ##e^{i\pi} + 1 = -1 + 1## hence ##e^{i\pi} + 1 = 0 ##

So sqrt of 1 has two roots 1 and -1 right? squaring the answer you got brings you back to the original ##e^{2\pi i} = 1##

@fresh_42 or @Mark44 can provide a better answer I think.

fresh_42
Mentor
• SACHIN KUMAR, Mr Davis 97 and jedishrfu
jedishrfu
Mentor
Thanks @fresh_42 I had forgotten about that insight by micro.

FactChecker
Science Advisor
Gold Member
Say I have ##e^{2\pi i n}##, where ##n## is an integer. Then it's clear that ##(e^{2\pi i})^n = 1^n = 1##.
However, what if replace ##n## with a rational number ##r##? It seems that by the same reasoning we should have that ##e^{2\pi i r} = (e^{2\pi i})^r = 1^r = 1##.
No more than the fact that ##sin(2\pi n) == 0## for all ##n\in N## implies ##sin(2\pi r) == 0## for all rational numbers ##r##
But what if ##r=1/2## for example? Then ##e^{2 \pi i (\frac{1}{2})} = e^{\pi i} = -1##. What am I doing wrong here?
The complex exponential function ##e^{i\theta}, \theta \in R## is very fundamental. I suggest that you study it in the context of complex analysis.