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

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.

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

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 $e^{i\pi/2}=i$.