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":

  1. 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 $$a^x = \lim_{n\to\infty} a^{x_n}.$$
  2. 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}} $?

  • $\begingroup$ what do you mean with $e$ in the argument of sin and cos? $\endgroup$ – Dr. Sonnhard Graubner Nov 11 '17 at 13:00
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    $\begingroup$ @OP: You can always use method 1. E.g., $4-\frac43+\frac45-\frac47+...$ converges to $\pi$. $\endgroup$ – user202729 Nov 11 '17 at 13:02
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    $\begingroup$ you can also have a look at WA, there are several formulas $\endgroup$ – Dr. Sonnhard Graubner Nov 11 '17 at 13:06
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    $\begingroup$ 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. $\endgroup$ – user202729 Nov 11 '17 at 13:13
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    $\begingroup$ @ArandomUserNameEG Wolfram Alpha. $\endgroup$ – user202729 Nov 11 '17 at 14:31

$$ (\cos e + i \sin e) ^ {\pi \over 2} $$ $$ = \left( e^ {i \times e} \right) ^ {\pi \over 2} $$ (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.

  • $\begingroup$ Now, that's an elegant solution, I didn't know you could use Euler's formula like that! $\endgroup$ – ArandomUserNameEG Nov 11 '17 at 13:55
  • $\begingroup$ You didn't know it? Good. Because the "law" $(e^a)^b = e^{ab}$ is in general false for complex $a,b$. $\endgroup$ – GEdgar Nov 11 '17 at 18:12

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