2
$\begingroup$

Calculate $e^{i \pi}$ and $ e^{i \pi /2}$ given that $2 \pi$ is the smallest real $>0$ such that $e^{2\pi i} = 1$.

I have done one part as:

Since $(e^{i \pi})^2 = e^{2\pi i} = 1$ we have $e^{i \pi} = +1$ or $-1$, but it can't be $+1$ as $2 \pi$ is the smallest real $>0$ such that $e^{2\pi i} = 1$ so $e^{i \pi} = -1$ .

But how to do the second part. Thank You.

$\endgroup$
  • $\begingroup$ If $e^{i\pi} = -1$ then $e^{i\pi/2} = (e^{i\pi})^{1/2} = ?$ $\endgroup$ – ÍgjøgnumMeg Dec 10 '16 at 1:51
  • 1
    $\begingroup$ @Jahambo99 There are always two solutions to $x^2=a$ $\endgroup$ – Simply Beautiful Art Dec 10 '16 at 1:54
  • 1
    $\begingroup$ it can be i or -i ..but how to determine among them?? $\endgroup$ – user8795 Dec 10 '16 at 1:58
1
$\begingroup$

If we have $\left(e^{i\pi/2}\right)^2=e^{\pi i}=-1$, then we know that $e^{i\pi/2}=\pm i$.

There is no algebraic way to determine whether $e^{i\pi/2}$ is $i$ or $-i$ since we are only using $\left(e^{i\pi/2}\right)^2=-1$ to define $e^{i\pi/2}$. The culprit here is the automorphism of $\mathbb{C}$ that swaps $i$ and $-i$ (see this answer).


To make $e^z$ complex differentiable, we must have $$ \begin{align} \lim_{n\to\infty}\frac{e^{i\pi/n}-1}{i\pi/n} &=\left.\frac{\mathrm{d}}{\mathrm{d}z}e^z\,\right|_{\,z=0}\\ &=1 \end{align} $$ Thus, $e^{i\pi/n}=1+\frac{i\pi}n+o\!\left(\frac1n\right)$ which places $e^{i\pi/n}$ into quadrant $1$.

Then, using trigonometry, and defining $e^{ix}=\lim\limits_{n\to\infty}\left(1+\frac{ix}n\right)^n$, we can use the method in this answer to show that $$ e^{ix}=\cos(x)+i\sin(x) $$ which shows that $e^{i\pi/2}=i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.