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Nov 03'24

Exercise

[math] \newcommand{\ex}[1]{\item } \newcommand{\sx}{\item} \newcommand{\x}{\sx} \newcommand{\sxlab}[1]{} \newcommand{\xlab}{\sxlab} \newcommand{\prov}[1] {\quad #1} \newcommand{\provx}[1] {\quad \mbox{#1}} \newcommand{\intext}[1]{\quad \mbox{#1} \quad} \newcommand{\R}{\mathrm{\bf R}} \newcommand{\Q}{\mathrm{\bf Q}} \newcommand{\Z}{\mathrm{\bf Z}} \newcommand{\C}{\mathrm{\bf C}} \newcommand{\dt}{\textbf} \newcommand{\goesto}{\rightarrow} \newcommand{\ddxof}[1]{\frac{d #1}{d x}} \newcommand{\ddx}{\frac{d}{dx}} \newcommand{\ddt}{\frac{d}{dt}} \newcommand{\dydx}{\ddxof y} \newcommand{\nxder}[3]{\frac{d^{#1}{#2}}{d{#3}^{#1}}} \newcommand{\deriv}[2]{\frac{d^{#1}{#2}}{dx^{#1}}} \newcommand{\dist}{\mathrm{distance}} \newcommand{\arccot}{\mathrm{arccot\:}} \newcommand{\arccsc}{\mathrm{arccsc\:}} \newcommand{\arcsec}{\mathrm{arcsec\:}} \newcommand{\arctanh}{\mathrm{arctanh\:}} \newcommand{\arcsinh}{\mathrm{arcsinh\:}} \newcommand{\arccosh}{\mathrm{arccosh\:}} \newcommand{\sech}{\mathrm{sech\:}} \newcommand{\csch}{\mathrm{csch\:}} \newcommand{\conj}[1]{\overline{#1}} \newcommand{\mathds}{\mathbb} [/math]

Every complex-valued function [math]f[/math] of a real variable determines two real-valued functions [math]f_1[/math] and [math]f_2[/math] of a real variable defined by

[[math]] f_1(x) = \mbox{real part of $f(x)$,} [[/math]]

[[math]] f_2(x) = \mbox{imaginary part of $f(x)$.} [[/math]]

Thus [math]f(x) = f_1(x) + if_2(x)[/math] for every [math]x[/math] in the domain of [math]f[/math]. We define the derivative [math]f^\prime[/math] by the formula

[[math]] f^\prime(x) = f_1^\prime(x) + i f_2^\prime(x) . [[/math]]

Applying this definition to the function [math]f(x) = e^{ix}[/math], show that

[[math]] \ddx e^{ix} = i e^{ix} . [[/math]]