Question

How do you show ${e^{it}} = \cos t + i\sin t$?

Answer 1

Here, in the given question, we need to show that ${e^{it}} = \cos t + i\sin t$. Here, $\cos$ and $\sin$ are trigonometric functions, $e$ = the base of natural logarithm, $i$ = imaginary unit and $t$ = angle in radians. First, we will express the exponential of a real number $x$, $\sin \theta$ and $\cos \theta$ as a sum of infinite series. As we know complex numbers are expressed as $z = a + ib$ so, we will express the exponential of a complex number $z$ in the same way as the exponential of a real number.In this, we will substitute, $z = it$ and simplify it further to get our answer.

Complete step by step answer:
The exponential of a real number $x$, written as ${e^x}$, is defined by sum of an infinite series, as follows
$\Rightarrow {e^x} = \sum\nolimits_{k = 0}^\infty {\left( {\dfrac{{{x^k}}}{{k!}}} \right)}$
On expansion, we get
$\Rightarrow {e^x} = 1 + x + \left( {\dfrac{{{x^2}}}{{2!}}} \right) + \left( {\dfrac{{{x^3}}}{{3!}}} \right) + \left( {\dfrac{{{x^4}}}{{4!}}} \right) + ....$

Also, $\cos \theta$ and $\sin \theta$ can be expressed as sum of infinite series as follows
$\Rightarrow \cos \left( \theta \right) = 1 - \left( {\dfrac{{{\theta ^2}}}{{2!}}} \right) + \left( {\dfrac{{{\theta ^4}}}{{4!}}} \right) + ....$
$\Rightarrow \sin \left( \theta \right) = \theta - \left( {\dfrac{{{\theta ^3}}}{{3!}}} \right) + \left( {\dfrac{{{\theta ^5}}}{{5!}}} \right) + ....$
(These are Taylor series).
The exponential of a complex number $z$ is written as ${e^z}$, and is defined in the same way as the exponential of a real number
$\Rightarrow {e^z} = \sum\nolimits_{k = 0}^\infty {\left( {\dfrac{{{z^k}}}{{k!}}} \right)}$
On expansion, we get
$\Rightarrow {e^z} = 1 + z + \left( {\dfrac{{{z^2}}}{{2!}}} \right) + \left( {\dfrac{{{z^3}}}{{3!}}} \right) + \left( {\dfrac{{{z^4}}}{{4!}}} \right) + ....$

Let $z = i.t$ in the previous relation
$\Rightarrow {e^{i.t}} = \sum\nolimits_{k = 0}^\infty {\left( {\dfrac{{{{\left( {i.t} \right)}^k}}}{{k!}}} \right)}$
On expansion, we get
$\Rightarrow {e^{i.t}} = 1 + \left( {i.t} \right) + \left( {\dfrac{{{{\left( {i.t} \right)}^2}}}{{2!}}} \right) + \left( {\dfrac{{{{\left( {i.t} \right)}^3}}}{{3!}}} \right) + \left( {\dfrac{{{{\left( {i.t} \right)}^4}}}{{4!}}} \right) + ....$
$\Rightarrow {e^{it}} = 1 + it + \left( {\dfrac{{{i^2}{t^2}}}{{2!}}} \right) + \left( {\dfrac{{i.{i^2}{t^3}}}{{3!}}} \right) + \left( {\dfrac{{{{\left( {{i^2}} \right)}^2}{t^4}}}{{4!}}} \right) + ....$

As we know ${i^2} = - 1$. So, on substituting this value we get,
$\Rightarrow {e^{it}} = 1 + it + \left( {\dfrac{{ - 1 \times {t^2}}}{{2!}}} \right) + \left( {\dfrac{{i \times - 1 \times {t^3}}}{{3!}}} \right) + \left( {\dfrac{{{{\left( { - 1} \right)}^2} \times {t^4}}}{{4!}}} \right) + ....$
On simplification, we get
$\Rightarrow {e^{it}} = 1 + it - \left( {\dfrac{{{t^2}}}{{2!}}} \right) - \left( {\dfrac{{i \times {t^3}}}{{3!}}} \right) + \left( {\dfrac{{{t^4}}}{{4!}}} \right) + ....$

On separating the terms which have $i$ and which are without $i$ and taking $i$ as a common term, we get
$\Rightarrow {e^{it}} = \left[ {1 - \left( {\dfrac{{{t^2}}}{{2!}}} \right) + \left( {\dfrac{{{t^4}}}{{4!}}} \right) + ....} \right] + i\left[ {t - \left( {\dfrac{{{t^3}}}{{3!}}} \right) + \left( {\dfrac{{{t^5}}}{{3!}}} \right) + ....} \right]$
As we know, $\cos \left( \theta \right) = 1 - \left( {\dfrac{{{\theta ^2}}}{{2!}}} \right) + \left( {\dfrac{{{\theta ^4}}}{{4!}}} \right) + ....$ and $\sin \left( \theta \right) = \theta - \left( {\dfrac{{{\theta ^3}}}{{3!}}} \right) + \left( {\dfrac{{{\theta ^5}}}{{5!}}} \right) + ....$. So, using this, we get
$\Rightarrow {e^{it}} = \cos t + i\sin t$
Thus the proof concluded.

Therefore, the identity ${e^{it}} = \cos t + i\sin t$ is known as Euler’s formula.

Note: Remember that $\sin e$ and $\cos ine$ functions are actually linear combinations of exponential functions with imaginary exponents. Also, the expression $\cos x + i\sin x$ is often referred to as $\operatorname{c} isx$. Note that, $\cos x$ is a even function (i.e., $\cos \left( { - x} \right) = + \cos \left( x \right)$) and the Taylor series of $\cos x$ has only even powers and $\sin x$ is an odd function (i.e., $\sin \left( { - x} \right) = - \sin \left( x \right)$) and the Taylor series of $\sin x$ has only odd powers. Remember that when $t = \pi$, ${e^{\pi i}} = - 1$ and when $t = 2\pi$, ${e^{2\pi i}} = 1$.