Show that $e^{iy} = 1 + iy + \frac{\mu_1 y^2}{2}$ for all $y \in \mathbb{R}$ with $|\mu_1| \leq 1$ How to show the expansions
\begin{gather*}
e^{iy} = 1 + iy + \frac{\mu_1 y^2}{2}\\
e^{iy} = 1 + iy - \frac{1}{2}y^2 + \frac{\mu_2 |y|^3}{3!}
\end{gather*}
where $y \in \mathbb{R}$ and $|\mu_1| \leq 1$ and $|\mu_2| \leq 1$ for all $y$?
I attempts to consider the remainder of Taylor expansion of one real-variable function: $e^{iy} = \cos y + i\sin y$, and their Taylor expansions with Lagrange remainders are
\begin{gather*}
\cos y = 1 - \frac{y^2}{2}\cos (\theta_1 y) y^2\\
\sin y = y - \frac{1}{2}\sin (\theta_2 y) y^2
\end{gather*}
So we may put $\mu_1 = -[\cos (\theta_1 y) + i \sin (\theta_2 y)]$. But here we only have $|\mu_1| \leq 2$, which does not satisfy the requirement...
Any help will be appreciated!

You made a mistake in your second expansion. The correct one is
\begin{align*}
\cos y = 1 - \cos(\theta_1y)\frac{y^2}{2}, &&
\sin y = y - \sin(\theta_2 y)\frac{y^3}{6}
\end{align*}
then
\begin{align*}
 e^{i y} &= \cos y + i \sin y \\
 & = 1 + i y - \cos(\theta_1y)\frac{y^2}{2}- i\sin(\theta_2 y)\frac{y^3}{6}
\end{align*}
which is closer to your answer ;)
