Can you raise a complex number to a transcendental exponent?

Question

For example, how do I find the value of:

$$ (\cos(e) + i\times\sin(e))^{\frac{\pi}{2}} $$

I've did a bit of searching and found this amazing question, and it gave me two ways "to raise a non-complex number to an irrational power":

You can pick a sequence of rational numbers $x_n$ converging to $x$ (i.e., $\lim\limits_{n\to\infty} x_n = x$) and define $$a^x = \lim_{n\to\infty} a^{x_n}.$$

You can use the exponential function $e^x$ (defined in many ways, say as $e^x = \lim_{n\to\infty} (1+\frac{x}n)^n$ or with a power series), and its inverse the logarithmic function that satifies $e^{\ln t} = t$ for all positive $t$, and since $a = e^{\ln a}$, define $$a^x = e^{x \ln a}.$$

The first way won't work with transcendental numbers such a $ \pi $, $ \frac{\pi}{2}$ and its family.

The second way will end up giving me $ \ln(i) $ in an exponent, which I'd rather not endure right now.

I tried De Moivre's, but I noticed that it only works for integer powers.

So how should I approach $(\cos(e) + i*\sin(e))^{\frac{\pi}{2}} $?

Euler constant, probably. Though it is very uncommon to use that. — user202729, Nov 11 '17 at 13:01
Sure. But if you do, beware that you have a branch cut like any other noninteger exponent. — Oscar Lanzi, Nov 11 '17 at 13:01
@OP: You can always use method 1. E.g., $4-\frac43+\frac45-\frac47+...$ converges to $\pi$. — user202729, Nov 11 '17 at 13:02
Euler formula is an extension of de Moivre theorem that also works for real powers. — user202729, Nov 11 '17 at 13:03
Also I suggest using \times for $\times$ because it is a more correct way to write multiplication (in this case); and there is a method to zoom math formulas (right click \ Math Settings \ Zoom Trigger \ Double-click then double-click at the formula), so including the "that tiny exponent" part is often unnecessary. — user202729, Nov 11 '17 at 13:13
Yes I meant the Euler constant,. @user202729: Thanks, I'll try doing that. — ArandomUserNameEG, Nov 11 '17 at 13:51
To be honest, this equation was from a trickier problem that I've simplified by using Euler's formula and arrived into this part, I couldn't figure out how to use Euler's formula to simplify it further, thanks. ----- I've used \times for x and deleted the unnecessary line. — ArandomUserNameEG, Nov 11 '17 at 13:55

user202729 · Accepted Answer · 2017-11-12T01:20:39.773

4

$$ (\cos e + i \sin e) ^ {\pi \over 2} $$ $$ = \left( e^ {i \times e} \right) ^ {\pi \over 2} $$ (by Euler's formula) $$ = e ^ {i \times \left({\pi \over 2} \times e\right)} $$ $$ = \cos \left({\pi \over 2} \times e\right) + i \sin \left({\pi \over 2} \times e\right) $$

Warning: As GEdgar wrote in the comment below, because of branch cut, $e^{x\times y}=\left(e^x\right)^y$ is not true for all complex $x$ and $y$. Wikipedia reference.

edited Nov 12 '17 at 01:20

answered Nov 11 '17 at 13:08

user202729

1,220

Now, that's an elegant solution, I didn't know you could use Euler's formula like that! – ArandomUserNameEG Nov 11 '17 at 13:55
1

You didn't know it? Good. Because the "law" $(e^a)^b = e^{ab}$ is in general false for complex $a,b$. – GEdgar Nov 11 '17 at 18:12

Can you raise a complex number to a transcendental exponent?

1 Answers1